OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Exercise 14 (F)

Question 1. Find the sum to
(i) 8 terms of \( 3 + 6 + 12 + ..... \)
(ii) 20 terms of \( 2 + 6 + 18 + ..... \)
(iii) 10 terms of \( 1 + \sqrt{3} + 3 + ..... \)
(iv) n terms of \( 3\frac{3}{8} + 2\frac{1}{4} + 1\frac{1}{2} + ..... \)
Answer:
(i) For the given series \( 3 + 6 + 12 + ...... \), which has 8 terms, the first term \( a = 3 \). The common ratio \( r = \frac{6}{3} = 2 \). Since \( r > 1 \), the sum \( S_n \) is calculated as:
\( S_n = \frac{a(r^n - 1)}{r-1} \)
\( S_8 = \frac{3(2^8 - 1)}{2-1} \)
\( S_8 = \frac{3(256 - 1)}{1} \)
\( S_8 = 3 \times 255 = 765 \)
(ii) For the given series \( 2 + 6 + 18 + ...... \), with 20 terms, the first term \( a = 2 \). The common ratio \( r = \frac{6}{2} = 3 \). Since \( r > 1 \), the sum \( S_n \) is:
\( S_n = \frac{a(r^n - 1)}{r-1} \)
\( S_{20} = \frac{2(3^{20} - 1)}{3-1} \)
\( S_{20} = \frac{2(3^{20} - 1)}{2} \)
\( S_{20} = 3^{20} - 1 \)
(iii) For the given series \( 1 + \sqrt{3} + 3 + ...... \), with 10 terms, the first term \( a = 1 \). The common ratio \( r = \frac{\sqrt{3}}{1} = \sqrt{3} \). Since \( \sqrt{3} \approx 1.732 > 1 \), the sum \( S_n \) is:
\( S_n = \frac{a(r^n - 1)}{r-1} \)
\( S_{10} = \frac{1((\sqrt{3})^{10} - 1)}{\sqrt{3} - 1} \)
\( S_{10} = \frac{3^5 - 1}{\sqrt{3} - 1} \)
Now, multiply by the conjugate \( (\sqrt{3} + 1) \) to simplify:
\( S_{10} = \frac{(3^5 - 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \)
\( S_{10} = \frac{(243 - 1)(\sqrt{3} + 1)}{3 - 1} \)
\( S_{10} = \frac{242(\sqrt{3} + 1)}{2} \)
\( S_{10} = 121(\sqrt{3} + 1) \)
(iv) For the given series \( 3\frac{3}{8} + 2\frac{1}{4} + 1\frac{1}{2} + ...... \), first convert the mixed fractions to improper fractions:
\( a = 3\frac{3}{8} = \frac{27}{8} \)
The second term is \( 2\frac{1}{4} = \frac{9}{4} \)
The common ratio \( r = \frac{\frac{9}{4}}{\frac{27}{8}} = \frac{9}{4} \times \frac{8}{27} = \frac{2}{3} \). Since \( r = \frac{2}{3} < 1 \), the sum \( S_n \) for n terms is:
\( S_n = \frac{a(1 - r^n)}{1-r} \)
\( S_n = \frac{\frac{27}{8}\left[1 - \left(\frac{2}{3}\right)^n\right]}{1 - \frac{2}{3}} \)
\( S_n = \frac{\frac{27}{8}\left[1 - \left(\frac{2}{3}\right)^n\right]}{\frac{1}{3}} \)
\( S_n = \frac{27}{8} \times 3 \left[1 - \left(\frac{2}{3}\right)^n\right] \)
\( S_n = \frac{81}{8}\left[1 - \left(\frac{2}{3}\right)^n\right] \)
In simple words: First, find the starting number (a) and how much it changes each time (r). If r is bigger than 1, use one formula for the sum; if r is smaller than 1, use a different formula. Just plug in the values and do the math carefully.

🎯 Exam Tip: Always identify the first term (a), the common ratio (r), and the number of terms (n) first. Remember to use the correct sum formula based on whether r is greater than or less than 1.

Question 2. Sum the following series to infinitely:
(i) \( 1 + \frac { 1 }{ 2 } + \frac { 1 }{ 4 } + \frac { 1 }{ 8 } + .... \)
(ii) \( 16, -8, 4 ...... \)
(iii) \( \sqrt{2} - \frac{1}{\sqrt{2}} + \frac{1}{2 \sqrt{2}} – \frac{1}{4 \sqrt{2}} + ...... \)
(iv) \( \sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{3 \sqrt{3}} ..... \)
Answer:
(i) For the series \( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .... \), the first term \( a = 1 \). The common ratio \( r = \frac{\frac{1}{2}}{1} = \frac{1}{2} \). Since \( |r| < 1 \), the sum to infinity \( S_\infty \) is:
\( S_\infty = \frac{a}{1-r} \)
\( S_\infty = \frac{1}{1 - \frac{1}{2}} \)
\( S_\infty = \frac{1}{\frac{1}{2}} = 2 \)
(ii) For the series \( 16 + (-8) + 4 + .... \), the first term \( a = 16 \). The common ratio \( r = \frac{-8}{16} = -\frac{1}{2} \). Since \( |r| < 1 \), the sum to infinity \( S_\infty \) is:
\( S_\infty = \frac{a}{1-r} \)
\( S_\infty = \frac{16}{1 - \left(-\frac{1}{2}\right)} \)
\( S_\infty = \frac{16}{1 + \frac{1}{2}} \)
\( S_\infty = \frac{16}{\frac{3}{2}} \)
\( S_\infty = 16 \times \frac{2}{3} = \frac{32}{3} \)
(iii) For the series \( \sqrt{2} - \frac{1}{\sqrt{2}} + \frac{1}{2 \sqrt{2}} – \frac{1}{4 \sqrt{2}} + ...... \), the first term \( a = \sqrt{2} \). The common ratio \( r = \frac{-\frac{1}{\sqrt{2}}}{\sqrt{2}} = -\frac{1}{2} \). Since \( |r| < 1 \), the sum to infinity \( S_\infty \) is:
\( S_\infty = \frac{a}{1-r} \)
\( S_\infty = \frac{\sqrt{2}}{1 - \left(-\frac{1}{2}\right)} \)
\( S_\infty = \frac{\sqrt{2}}{1 + \frac{1}{2}} \)
\( S_\infty = \frac{\sqrt{2}}{\frac{3}{2}} \)
\( S_\infty = \frac{2\sqrt{2}}{3} \)
(iv) For the series \( \sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{3 \sqrt{3}} ..... \), the first term \( a = \sqrt{3} \). The common ratio \( r = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{3} \). Since \( |r| < 1 \), the sum to infinity \( S_\infty \) is:
\( S_\infty = \frac{a}{1-r} \)
\( S_\infty = \frac{\sqrt{3}}{1 - \frac{1}{3}} \)
\( S_\infty = \frac{\sqrt{3}}{\frac{2}{3}} \)
\( S_\infty = \frac{3\sqrt{3}}{2} \)
In simple words: To find the sum of a series that goes on forever (infinite series), you need to know the first number and how it changes by multiplication. If the common ratio is between -1 and 1, use the special formula \( a/(1-r) \).

🎯 Exam Tip: Remember that a geometric series only has a sum to infinity if the absolute value of the common ratio \( |r| \) is less than 1. If \( |r| \ge 1 \), the series diverges and has no finite sum.

Question 3. Find the sum of a geometric series in which a = 16, r = \( \frac{1}{4} \), l = \( \frac{1}{64} \)
Answer:
Given: First term \( a = 16 \), common ratio \( r = \frac{1}{4} \), and last term \( l = \frac{1}{64} \).
First, we need to find the number of terms \( n \) using the formula for the nth term of a G.P., \( T_n = ar^{n-1} \).
We are given \( T_n = l = \frac{1}{64} \).
\( \frac{1}{64} = 16 \times \left(\frac{1}{4}\right)^{n-1} \)
Divide both sides by 16:
\( \frac{1}{64 \times 16} = \left(\frac{1}{4}\right)^{n-1} \)
\( \frac{1}{1024} = \left(\frac{1}{4}\right)^{n-1} \)
We know that \( 4^5 = 1024 \), so \( \frac{1}{1024} = \left(\frac{1}{4}\right)^5 \).
\( \left(\frac{1}{4}\right)^5 = \left(\frac{1}{4}\right)^{n-1} \)
This means the exponents must be equal:
\( 5 = n-1 \)
\( n = 6 \)
Now, we find the sum of the 6 terms using the formula for \( S_n \) when \( r < 1 \):
\( S_n = \frac{a(1 - r^n)}{1-r} \)
\( S_6 = \frac{16\left(1 - \left(\frac{1}{4}\right)^6\right)}{1 - \frac{1}{4}} \)
\( S_6 = \frac{16\left(1 - \frac{1}{4096}\right)}{\frac{3}{4}} \)
\( S_6 = \frac{16\left(\frac{4096 - 1}{4096}\right)}{\frac{3}{4}} \)
\( S_6 = \frac{16 \times \frac{4095}{4096}}{\frac{3}{4}} \)
\( S_6 = \frac{16 \times 4095}{4096} \times \frac{4}{3} \)
\( S_6 = \frac{4095}{256} \times \frac{4}{3} \)
\( S_6 = \frac{1365}{64} \)
In simple words: First, figure out how many numbers are in the series using the last number given. Once you know the total count of numbers, use the sum formula for a geometric series to add them all up.

🎯 Exam Tip: When given the first term, common ratio, and last term, always find the number of terms (n) first. Then, use the appropriate sum formula based on the value of r.

Question 4. Find the sum \( 81 - 27 + 9 - ... - \frac{1}{27} \)
Answer:
Given the series \( 81 - 27 + 9 - ... - \frac{1}{27} \).
This is a geometric progression (G.P.).
The first term \( a = 81 \).
The common ratio \( r = \frac{-27}{81} = -\frac{1}{3} \).
The last term \( l = -\frac{1}{27} \).
First, we need to find the number of terms \( n \) using the formula \( T_n = ar^{n-1} \):
\( -\frac{1}{27} = 81 \times \left(-\frac{1}{3}\right)^{n-1} \)
Divide both sides by 81:
\( -\frac{1}{27 \times 81} = \left(-\frac{1}{3}\right)^{n-1} \)
\( -\frac{1}{3^3 \times 3^4} = \left(-\frac{1}{3}\right)^{n-1} \)
\( -\frac{1}{3^7} = \left(-\frac{1}{3}\right)^{n-1} \)
\( \left(-\frac{1}{3}\right)^7 = \left(-\frac{1}{3}\right)^{n-1} \)
Equating the exponents, we get:
\( 7 = n-1 \)
\( n = 8 \)
Now, we find the sum of the 8 terms using the formula for \( S_n \) when \( r < 1 \):
\( S_n = \frac{a(1 - r^n)}{1-r} \)
\( S_8 = \frac{81\left(1 - \left(-\frac{1}{3}\right)^8\right)}{1 - \left(-\frac{1}{3}\right)} \)
\( S_8 = \frac{81\left(1 - \frac{1}{3^8}\right)}{1 + \frac{1}{3}} \)
\( S_8 = \frac{81\left(1 - \frac{1}{6561}\right)}{\frac{4}{3}} \)
\( S_8 = \frac{81\left(\frac{6561 - 1}{6561}\right)}{\frac{4}{3}} \)
\( S_8 = \frac{81 \times \frac{6560}{6561}}{\frac{4}{3}} \)
\( S_8 = \frac{6560}{81} \times \frac{3}{4} \)
\( S_8 = \frac{1640}{27} \)
In simple words: This series goes down in value by a fraction each time. First, find out how many numbers are in the series. Then, use the special sum formula for a geometric series where the change is less than one.

🎯 Exam Tip: Pay close attention to the sign of the common ratio. A negative ratio means the terms will alternate in sign, which affects the sum calculation.

Question 5. The first three terms of a G.P. are x, x + 3, x + 9. Find the value of x and the sum of first eight terms.
Answer:
Given that \( x, x + 3, \) and \( x + 9 \) are the first three terms of a Geometric Progression (G.P.).
In a G.P., the ratio of consecutive terms is constant. So, the common ratio \( r \) can be found by:
\( \frac{x+3}{x} = \frac{x+9}{x+3} \)
Cross-multiply to solve for x:
\( (x+3)^2 = x(x+9) \)
Expand both sides:
\( x^2 + 6x + 9 = x^2 + 9x \)
Subtract \( x^2 \) from both sides:
\( 6x + 9 = 9x \)
Subtract \( 6x \) from both sides:
\( 9 = 3x \)
Divide by 3:
\( x = 3 \)
Now substitute \( x=3 \) back into the terms to find the G.P.:
First term \( a = x = 3 \)
Second term \( = x + 3 = 3 + 3 = 6 \)
Third term \( = x + 9 = 3 + 9 = 12 \)
The G.P. is \( 3, 6, 12, ... \).
The common ratio \( r = \frac{6}{3} = 2 \). Since \( r > 1 \), we use the sum formula \( S_n = \frac{a(r^n - 1)}{r-1} \).
We need to find the sum of the first eight terms, so \( n = 8 \).
\( S_8 = \frac{3(2^8 - 1)}{2-1} \)
\( S_8 = \frac{3(256 - 1)}{1} \)
\( S_8 = 3 \times 255 \)
\( S_8 = 765 \)
In simple words: If three numbers are in a geometric progression, the middle number squared is equal to the first number multiplied by the third number. Use this rule to find 'x'. Once you have 'x', find the first term and the common ratio, then calculate the sum of the first eight terms.

🎯 Exam Tip: A key property of a G.P. is that the square of any term is equal to the product of its preceding and succeeding terms (e.g., \( b^2 = ac \) for a, b, c in G.P.). Use this to solve for unknowns like x.

Question 6. Of how many terms is \( \frac{55}{72} \), the sum of the series \( \frac{2}{9} – \frac{1}{3} + ... \)
Answer:
Given the series \( \frac{2}{9} – \frac{1}{3} + ... \), and the sum \( S_n = \frac{55}{72} \).
This is a geometric progression.
The first term \( a = \frac{2}{9} \).
The common ratio \( r = \frac{-\frac{1}{3}}{\frac{2}{9}} = -\frac{1}{3} \times \frac{9}{2} = -\frac{3}{2} \).
Since \( |r| = \left|-\frac{3}{2}\right| = \frac{3}{2} > 1 \), we use the sum formula \( S_n = \frac{a(r^n - 1)}{r-1} \).
We are given \( S_n = \frac{55}{72} \).
\( \frac{55}{72} = \frac{\frac{2}{9}\left(\left(-\frac{3}{2}\right)^n - 1\right)}{-\frac{3}{2} - 1} \)
Simplify the denominator:
\( -\frac{3}{2} - 1 = -\frac{3}{2} - \frac{2}{2} = -\frac{5}{2} \)
So, the equation becomes:
\( \frac{55}{72} = \frac{\frac{2}{9}\left(\left(-\frac{3}{2}\right)^n - 1\right)}{-\frac{5}{2}} \)
\( \frac{55}{72} = \frac{2}{9} \times \left(-\frac{2}{5}\right) \left(\left(-\frac{3}{2}\right)^n - 1\right) \)
\( \frac{55}{72} = -\frac{4}{45} \left(\left(-\frac{3}{2}\right)^n - 1\right) \)
Multiply both sides by \( -\frac{45}{4} \):
\( \frac{55}{72} \times \left(-\frac{45}{4}\right) = \left(-\frac{3}{2}\right)^n - 1 \)
\( -\frac{55 \times 5}{8 \times 4} = \left(-\frac{3}{2}\right)^n - 1 \)
\( -\frac{275}{32} = \left(-\frac{3}{2}\right)^n - 1 \)
Add 1 to both sides:
\( 1 - \frac{275}{32} = \left(-\frac{3}{2}\right)^n \)
\( \frac{32 - 275}{32} = \left(-\frac{3}{2}\right)^n \)
\( -\frac{243}{32} = \left(-\frac{3}{2}\right)^n \)
We know that \( 243 = 3^5 \) and \( 32 = 2^5 \).
So, \( -\frac{243}{32} = -\left(\frac{3}{2}\right)^5 = \left(-\frac{3}{2}\right)^5 \).
Therefore,
\( \left(-\frac{3}{2}\right)^5 = \left(-\frac{3}{2}\right)^n \)
Equating the exponents:
\( n = 5 \)
In simple words: We are given the total sum of a series and the first few numbers in that series. First, find out the starting number and how much it changes each time. Then, use the sum formula for geometric progressions and work backward to find how many terms were added up to get the given total sum.

🎯 Exam Tip: When the common ratio is negative, remember to handle the signs carefully, especially when raising it to a power. An odd power will keep the negative sign, while an even power will make it positive.

Question 7. The second term of a G.P. is 2 and the sum of infinite terms is 8. Find the first term.
Answer:
Let \( a \) be the first term and \( r \) be the common ratio of the G.P.
Given that the second term \( T_2 = 2 \). The formula for the nth term is \( T_n = ar^{n-1} \).
So, for \( T_2 \):
\( ar = 2 \quad ...(1) \)
Given that the sum of infinite terms \( S_\infty = 8 \). The formula for the sum to infinity is \( S_\infty = \frac{a}{1-r} \).
So:
\( \frac{a}{1-r} = 8 \quad ...(2) \)
From equation (1), we can express \( r \) in terms of \( a \):
\( r = \frac{2}{a} \)
Substitute this value of \( r \) into equation (2):
\( \frac{a}{1 - \frac{2}{a}} = 8 \)
Simplify the denominator:
\( \frac{a}{\frac{a-2}{a}} = 8 \)
\( \frac{a^2}{a-2} = 8 \)
Multiply both sides by \( (a-2) \):
\( a^2 = 8(a-2) \)
\( a^2 = 8a - 16 \)
Rearrange into a quadratic equation:
\( a^2 - 8a + 16 = 0 \)
This is a perfect square trinomial: \( (a-4)^2 \).
\( (a-4)^2 = 0 \)
Take the square root of both sides:
\( a-4 = 0 \)
\( a = 4 \)
Thus, the first term of the G.P. is 4.
We can also find \( r \) using \( r = \frac{2}{a} = \frac{2}{4} = \frac{1}{2} \). Since \( |r| = \frac{1}{2} < 1 \), the sum to infinity exists.
In simple words: We know the second number in a special series and what all the numbers in that series add up to if it goes on forever. We use these two facts to make equations and solve them to find the very first number of the series.

🎯 Exam Tip: Remember to check that the common ratio \( |r| < 1 \) after finding it, as this condition is necessary for the sum to infinity to exist. If it doesn't meet this condition, there's likely an error in the calculation.

Question 8.
(i) Find the value of 0.234 regarding it as a geometric series.
(ii) Evaluate : (a) \( 0.9\overline{7} \) (b) \( 0.\overline{45} \) (c) \( 0.23\overline{45} \)
(iii) Find a rational number which when expressed as a decimal will have \( 1.2\overline{56} \) as its expansion.
Answer:
(i) To find the value of \( 0.234 \) as a geometric series, it is implied that the digit 4 repeats indefinitely, meaning \( 0.23\overline{4} \). Let's interpret \( 0.234 \) as \( 0.23\overline{4} \).
\( 0.23\overline{4} = 0.234444... \)
We can write this as a sum:
\( 0.23 + 0.004 + 0.0004 + 0.00004 + ... \)
\( = \frac{23}{100} + \frac{4}{1000} + \frac{4}{10000} + \frac{4}{100000} + ... \)
The repeating part \( \left(\frac{4}{1000} + \frac{4}{10000} + ...\right) \) is an infinite G.P. with first term \( a = \frac{4}{1000} \) and common ratio \( r = \frac{1}{10} \).
The sum of this infinite G.P. is \( S_\infty = \frac{a}{1-r} \).
\( S_\infty = \frac{\frac{4}{1000}}{1 - \frac{1}{10}} = \frac{\frac{4}{1000}}{\frac{9}{10}} = \frac{4}{1000} \times \frac{10}{9} = \frac{4}{900} = \frac{1}{225} \)
Now, add the non-repeating part:
\( 0.23\overline{4} = \frac{23}{100} + \frac{1}{225} \)
Find a common denominator, which is 900.
\( = \frac{23 \times 9}{100 \times 9} + \frac{1 \times 4}{225 \times 4} = \frac{207}{900} + \frac{4}{900} = \frac{211}{900} \)
However, the OCR's provided solution for 0.234 interprets it as 0.2343434... which means 34 is the repetend. Let's follow that interpretation.
\( 0.23\overline{34} = 0.2 + 0.034 + 0.00034 + 0.0000034 + ... \)
\( = \frac{2}{10} + \left(\frac{34}{1000} + \frac{34}{100000} + \frac{34}{10000000} + ...\right) \)
The part in the parenthesis is an infinite G.P. with first term \( a = \frac{34}{1000} \) and common ratio \( r = \frac{1}{100} \).
Sum of infinite G.P.: \( S_\infty = \frac{a}{1-r} = \frac{\frac{34}{1000}}{1-\frac{1}{100}} = \frac{\frac{34}{1000}}{\frac{99}{100}} = \frac{34}{1000} \times \frac{100}{99} = \frac{34}{990} \)
Now add the non-repeating part \( \frac{2}{10} \):
\( 0.2\overline{34} = \frac{2}{10} + \frac{34}{990} \)
\( = \frac{2 \times 99}{10 \times 99} + \frac{34}{990} = \frac{198}{990} + \frac{34}{990} = \frac{232}{990} \)
Simplify the fraction by dividing by 2:
\( = \frac{116}{495} \)
(ii) Evaluate:
(a) \( 0.9\overline{7} \)
\( 0.9\overline{7} = 0.9 + 0.07 + 0.007 + 0.0007 + ... \)
\( = \frac{9}{10} + \left(\frac{7}{100} + \frac{7}{1000} + \frac{7}{10000} + ...\right) \)
The repeating part is an infinite G.P. with first term \( a = \frac{7}{100} \) and common ratio \( r = \frac{1}{10} \).
Sum of infinite G.P.: \( S_\infty = \frac{a}{1-r} = \frac{\frac{7}{100}}{1-\frac{1}{10}} = \frac{\frac{7}{100}}{\frac{9}{10}} = \frac{7}{100} \times \frac{10}{9} = \frac{7}{90} \)
Add the non-repeating part:
\( 0.9\overline{7} = \frac{9}{10} + \frac{7}{90} \)
\( = \frac{9 \times 9}{10 \times 9} + \frac{7}{90} = \frac{81}{90} + \frac{7}{90} = \frac{88}{90} \)
Simplify the fraction by dividing by 2:
\( = \frac{44}{45} \)
(b) \( 0.\overline{45} \)
\( 0.\overline{45} = 0.45 + 0.0045 + 0.000045 + ... \)
This is an infinite G.P. with first term \( a = \frac{45}{100} \) and common ratio \( r = \frac{1}{100} \).
Sum of infinite G.P.: \( S_\infty = \frac{a}{1-r} = \frac{\frac{45}{100}}{1-\frac{1}{100}} = \frac{\frac{45}{100}}{\frac{99}{100}} = \frac{45}{100} \times \frac{100}{99} = \frac{45}{99} \)
Simplify the fraction by dividing by 9:
\( = \frac{5}{11} \)
(c) \( 0.23\overline{45} \)
\( 0.23\overline{45} = 0.23 + 0.0045 + 0.000045 + 0.00000045 + ... \)
\( = \frac{23}{100} + \left(\frac{45}{10000} + \frac{45}{1000000} + ...\right) \)
The repeating part is an infinite G.P. with first term \( a = \frac{45}{10000} \) and common ratio \( r = \frac{1}{100} \).
Sum of infinite G.P.: \( S_\infty = \frac{a}{1-r} = \frac{\frac{45}{10000}}{1-\frac{1}{100}} = \frac{\frac{45}{10000}}{\frac{99}{100}} = \frac{45}{10000} \times \frac{100}{99} = \frac{45}{9900} \)
Simplify \( \frac{45}{9900} \) by dividing by 45:
\( = \frac{1}{220} \)
Add the non-repeating part:
\( 0.23\overline{45} = \frac{23}{100} + \frac{1}{220} \)
Find a common denominator, which is 1100.
\( = \frac{23 \times 11}{100 \times 11} + \frac{1 \times 5}{220 \times 5} = \frac{253}{1100} + \frac{5}{1100} = \frac{258}{1100} \)
Simplify the fraction by dividing by 2:
\( = \frac{129}{550} \)
(iii) Find a rational number which when expressed as a decimal will have \( 1.2\overline{56} \) as its expansion.
\( 1.2\overline{56} = 1.2 + 0.056 + 0.00056 + 0.0000056 + ... \)
\( = \frac{12}{10} + \left(\frac{56}{1000} + \frac{56}{100000} + \frac{56}{10000000} + ...\right) \)
The repeating part is an infinite G.P. with first term \( a = \frac{56}{1000} \) and common ratio \( r = \frac{1}{100} \).
Sum of infinite G.P.: \( S_\infty = \frac{a}{1-r} = \frac{\frac{56}{1000}}{1-\frac{1}{100}} = \frac{\frac{56}{1000}}{\frac{99}{100}} = \frac{56}{1000} \times \frac{100}{99} = \frac{56}{990} \)
Add the non-repeating part:
\( 1.2\overline{56} = \frac{12}{10} + \frac{56}{990} \)
Find a common denominator, which is 990.
\( = \frac{12 \times 99}{10 \times 99} + \frac{56}{990} = \frac{1188}{990} + \frac{56}{990} = \frac{1244}{990} \)
Simplify the fraction by dividing by 2:
\( = \frac{622}{495} \)
In simple words: To change a repeating decimal into a fraction, first separate the non-repeating part from the repeating part. The repeating part can be written as an infinite geometric series. Then, use the formula for the sum to infinity, and finally add it to the non-repeating part to get the full fraction.

🎯 Exam Tip: When converting repeating decimals to fractions, correctly identify the first term (a) and common ratio (r) of the geometric series formed by the repeating digits. Be careful with the powers of 10 in the denominators based on the position of the decimal point and the number of repeating digits.

Question 9. If \( a + b + ... + l \) is a G.P., prove that its sum is \( \frac{bl-a^2}{b-a} \).
Answer:
Given a Geometric Progression (G.P.) with terms \( a, b, ..., l \).
The first term is \( a \).
The common ratio \( r = \frac{b}{a} \).
The last term is \( l \).
We know the formula for the sum of \( n \) terms of a G.P. is \( S_n = \frac{a(r^n - 1)}{r-1} \).
We also know that the nth term \( T_n = ar^{n-1} \). So, the last term \( l = ar^{n-1} \).
From this, we can write \( ar^n = rl \).
Now, rewrite the sum formula:
\( S_n = \frac{ar^n - a}{r-1} \)
Substitute \( ar^n = rl \):
\( S_n = \frac{rl - a}{r-1} \)
Now, substitute \( r = \frac{b}{a} \) into this equation:
\( S_n = \frac{\left(\frac{b}{a}\right)l - a}{\frac{b}{a}-1} \)
Simplify the numerator and denominator:
\( S_n = \frac{\frac{bl}{a} - a}{\frac{b-a}{a}} \)
\( S_n = \frac{\frac{bl - a^2}{a}}{\frac{b-a}{a}} \)
The \( a \) in the denominator of the numerator and the denominator of the main fraction cancel out:
\( S_n = \frac{bl - a^2}{b-a} \)
Thus, it is proven that the sum of the G.P. is \( \frac{bl-a^2}{b-a} \).
In simple words: For a geometric series, if you know the first term, the second term, and the last term, there's a quick way to find the total sum. This formula connects these three values to give you the sum without needing to know how many terms are there.

🎯 Exam Tip: This is an alternative sum formula for a G.P. that is very useful when the number of terms (n) is not explicitly given but the last term (l) is. Make sure to derive it once to understand its components.

Question 10. The nth term of a geometrical progression is \( \frac{2^{2n-1}}{3} \) for all values of the first three terms and calculate the sum of the first 10 terms, correct to 3 significant figures.
Answer:
Given the nth term of a G.P. as \( T_n = \frac{2^{2n-1}}{3} \).
Let's find the first three terms:
For \( n=1 \), \( T_1 = \frac{2^{2(1)-1}}{3} = \frac{2^{2-1}}{3} = \frac{2^1}{3} = \frac{2}{3} \). So, the first term \( a = \frac{2}{3} \).
For \( n=2 \), \( T_2 = \frac{2^{2(2)-1}}{3} = \frac{2^{4-1}}{3} = \frac{2^3}{3} = \frac{8}{3} \).
For \( n=3 \), \( T_3 = \frac{2^{2(3)-1}}{3} = \frac{2^{6-1}}{3} = \frac{2^5}{3} = \frac{32}{3} \).
Now, find the common ratio \( r \):
\( r = \frac{T_2}{T_1} = \frac{\frac{8}{3}}{\frac{2}{3}} = \frac{8}{2} = 4 \).
(As a check, \( r = \frac{T_3}{T_2} = \frac{\frac{32}{3}}{\frac{8}{3}} = \frac{32}{8} = 4 \). The common ratio is consistent.)
Since \( r = 4 > 1 \), we use the sum formula \( S_n = \frac{a(r^n - 1)}{r-1} \).
We need to find the sum of the first 10 terms, so \( n=10 \).
\( S_{10} = \frac{\frac{2}{3}(4^{10} - 1)}{4-1} \)
\( S_{10} = \frac{\frac{2}{3}(4^{10} - 1)}{3} \)
\( S_{10} = \frac{2}{9}(4^{10} - 1) \)
Calculate \( 4^{10} \):
\( 4^{10} = (2^2)^{10} = 2^{20} = 1,048,576 \).
\( S_{10} = \frac{2}{9}(1,048,576 - 1) \)
\( S_{10} = \frac{2}{9}(1,048,575) \)
\( S_{10} = \frac{2,097,150}{9} \)
\( S_{10} = 233,016.666... \)
Rounding to 3 significant figures:
\( S_{10} = 233,000 \)
In simple words: First, use the given rule for the nth term to find the first few numbers in the series. From these, find the starting number and how much it grows by. Then, use the sum formula to add up the first 10 numbers and round the final answer to three important digits.

🎯 Exam Tip: Always make sure to calculate the common ratio from the first two terms. If the nth term formula is given, test it for \( n=1, 2, 3 \) to confirm it's a G.P. before proceeding. Remember to apply rounding rules at the very end of the calculation.

Question 11. A geometrical progression of positive terms and an arithmetical progression have the same first term. The sum of their first terms is 1, the sum of their second terms is \( \frac{1}{2} \) and the sum of their third terms is 2. Calculate the sum of their fourth terms.
Answer:
Let the first term for both A.P. and G.P. be \( a \).
Let the common difference of the A.P. be \( d \). The A.P. terms are \( a, a+d, a+2d, ... \).
Let the common ratio of the G.P. be \( r \). The G.P. terms are \( a, ar, ar^2, ... \).
Given that the terms of the G.P. are positive, which means \( a > 0 \) and \( r > 0 \).

1. Sum of their first terms is 1:
\( a + a = 1 \)
\( 2a = 1 \)
\( a = \frac{1}{2} \)

2. Sum of their second terms is \( \frac{1}{2} \):
\( (a+d) + ar = \frac{1}{2} \)
Substitute \( a = \frac{1}{2} \):
\( \left(\frac{1}{2} + d\right) + \frac{1}{2}r = \frac{1}{2} \)
Subtract \( \frac{1}{2} \) from both sides:
\( d + \frac{1}{2}r = 0 \)
Multiply by 2:
\( 2d + r = 0 \quad ...(1) \)

3. Sum of their third terms is 2:
\( (a+2d) + ar^2 = 2 \)
Substitute \( a = \frac{1}{2} \):
\( \left(\frac{1}{2} + 2d\right) + \frac{1}{2}r^2 = 2 \)
Multiply by 2:
\( 1 + 4d + r^2 = 4 \)
\( 4d + r^2 = 3 \quad ...(2) \)

From equation (1), \( r = -2d \).
Substitute \( r = -2d \) into equation (2):
\( 4d + (-2d)^2 = 3 \)
\( 4d + 4d^2 = 3 \)
Rearrange into a quadratic equation:
\( 4d^2 + 4d - 3 = 0 \)
Factor the quadratic equation:
\( 4d^2 + 6d - 2d - 3 = 0 \)
\( 2d(2d+3) - 1(2d+3) = 0 \)
\( (2d-1)(2d+3) = 0 \)
This gives two possible values for \( d \):
\( 2d - 1 = 0 \implies d = \frac{1}{2} \)
\( 2d + 3 = 0 \implies d = -\frac{3}{2} \)

Now find the corresponding values for \( r \) using \( r = -2d \):
If \( d = \frac{1}{2} \), then \( r = -2\left(\frac{1}{2}\right) = -1 \).
If \( d = -\frac{3}{2} \), then \( r = -2\left(-\frac{3}{2}\right) = 3 \).

Since the G.P. has positive terms, the common ratio \( r \) must be positive. Therefore, \( r = -1 \) is not possible. We must choose \( r=3 \).
If \( r = 3 \), then \( d = -\frac{3}{2} \).

So, we have: \( a = \frac{1}{2} \), \( d = -\frac{3}{2} \), and \( r = 3 \).

We need to calculate the sum of their fourth terms:
Fourth term of A.P. \( = a + 3d \)
Fourth term of G.P. \( = ar^3 \)
Sum of fourth terms \( = (a + 3d) + ar^3 \)
Substitute the values:
\( = \left(\frac{1}{2} + 3\left(-\frac{3}{2}\right)\right) + \frac{1}{2}(3)^3 \)
\( = \left(\frac{1}{2} - \frac{9}{2}\right) + \frac{1}{2}(27) \)
\( = \frac{1-9}{2} + \frac{27}{2} \)
\( = -\frac{8}{2} + \frac{27}{2} \)
\( = -4 + \frac{27}{2} \)
\( = \frac{-8 + 27}{2} = \frac{19}{2} \)
In simple words: We have two types of number patterns, one adding numbers and one multiplying them, both starting with the same number. We're given rules about what their first, second, and third numbers add up to. We use these rules to find the starting number, the adding amount, and the multiplying amount for each pattern. Finally, we calculate what their fourth numbers would add up to.

🎯 Exam Tip: When dealing with multiple sequences, clearly define variables for each (e.g., \( d \) for A.P., \( r \) for G.P.). Pay close attention to conditions like "positive terms" as they help eliminate incorrect solutions for \( r \).

Question 12. In a geometric progression, the third term exceeds the second by 6 and the second exceeds the first by 9. Find
(i) the first term, (ii) the common ratio and (iii) the sum of the first ten terms.
Answer:
Let \( a \) be the first term and \( r \) be the common ratio of the G.P.
The terms of the G.P. are \( a, ar, ar^2, ar^3, ... \).

Given conditions:
1. The third term exceeds the second by 6:
\( T_3 = T_2 + 6 \)
\( ar^2 = ar + 6 \)
\( ar^2 - ar = 6 \quad ...(1) \)

2. The second term exceeds the first by 9:
\( T_2 = T_1 + 9 \)
\( ar = a + 9 \)
\( ar - a = 9 \quad ...(2) \)

From (1), factor out \( ar \): \( ar(r-1) = 6 \)
From (2), factor out \( a \): \( a(r-1) = 9 \)

Divide equation (1) by equation (2):
\( \frac{ar(r-1)}{a(r-1)} = \frac{6}{9} \)
\( r = \frac{2}{3} \)
This is the common ratio.

(i) To find the first term, substitute \( r = \frac{2}{3} \) into equation (2):
\( a\left(\frac{2}{3} - 1\right) = 9 \)
\( a\left(-\frac{1}{3}\right) = 9 \)
\( a = -27 \)
So, the first term is \( -27 \).

(ii) The common ratio is \( \frac{2}{3} \).

(iii) To find the sum of the first ten terms, \( S_{10} \).
Since \( a = -27 \), \( r = \frac{2}{3} \), and \( |r| < 1 \), use the sum formula \( S_n = \frac{a(1 - r^n)}{1-r} \).
\( S_{10} = \frac{-27\left(1 - \left(\frac{2}{3}\right)^{10}\right)}{1 - \frac{2}{3}} \)
\( S_{10} = \frac{-27\left(1 - \frac{2^{10}}{3^{10}}\right)}{\frac{1}{3}} \)
\( S_{10} = -27 \times 3 \left(1 - \frac{1024}{59049}\right) \)
\( S_{10} = -81 \left(\frac{59049 - 1024}{59049}\right) \)
\( S_{10} = -81 \left(\frac{58025}{59049}\right) \)
Since \( 59049 = 81 \times 729 \):
\( S_{10} = -\frac{58025}{729} \)
In simple words: We're given clues about how the terms in a geometric series relate to each other. We use these clues to set up equations. Solving these equations helps us find the starting number, the factor by which numbers multiply, and then the sum of the first ten numbers in that series.

🎯 Exam Tip: Always write down the given information as equations using the standard G.P. formulas \( T_n = ar^{n-1} \) and \( S_n = \frac{a(1-r^n)}{1-r} \) (or \( \frac{a(r^n-1)}{r-1} \) if \( r>1 \)). Solving these simultaneous equations is the key to finding \( a \) and \( r \).

Question 13. In an infinite geometric progression, the sum of first two terms is 6 and every term is four times the sum of all the terms that follow it. Find :
(i) the geometric progression and
(ii) its sum to infinity.
Answer:
Let \( a \) be the first term and \( r \) be the common ratio of the infinite G.P.
(i) Find the geometric progression:

1. The sum of the first two terms is 6:
\( T_1 + T_2 = 6 \)
\( a + ar = 6 \)
\( a(1+r) = 6 \quad ...(1) \)

2. Every term is four times the sum of all terms that follow it.
Let's consider the \( n^{th} \) term, \( T_n = ar^{n-1} \).
The terms that follow \( T_n \) are \( ar^n, ar^{n+1}, ar^{n+2}, ... \). This is an infinite G.P. starting with \( ar^n \) and having the same common ratio \( r \).
The sum of terms following \( T_n \) is \( S_{following} = \frac{ar^n}{1-r} \).
According to the condition:
\( T_n = 4 \times S_{following} \)
\( ar^{n-1} = 4 \times \frac{ar^n}{1-r} \)
Divide both sides by \( ar^{n-1} \) (assuming \( a \neq 0 \) and \( r \neq 0 \)):
\( 1 = 4 \times \frac{r}{1-r} \)
Multiply both sides by \( (1-r) \):
\( 1-r = 4r \)
\( 1 = 5r \)
\( r = \frac{1}{5} \)
Now substitute \( r = \frac{1}{5} \) into equation (1) to find \( a \):
\( a\left(1+\frac{1}{5}\right) = 6 \)
\( a\left(\frac{6}{5}\right) = 6 \)
\( a = 5 \)
So, the first term is 5 and the common ratio is \( \frac{1}{5} \).
The geometric progression is \( a, ar, ar^2, ... \), which is:
\( 5, 5\left(\frac{1}{5}\right), 5\left(\frac{1}{5}\right)^2, ... \)
\( 5, 1, \frac{1}{5}, ... \)

(ii) Find its sum to infinity:
For an infinite G.P., the sum to infinity \( S_\infty = \frac{a}{1-r} \).
Using \( a=5 \) and \( r=\frac{1}{5} \):
\( S_\infty = \frac{5}{1 - \frac{1}{5}} \)
\( S_\infty = \frac{5}{\frac{4}{5}} \)
\( S_\infty = 5 \times \frac{5}{4} = \frac{25}{4} \)
In simple words: We're given two clues about a special series that never ends: what its first two numbers add up to, and a rule about how each number relates to all the numbers that come after it. We use these clues to find the starting number and the multiplying factor for the series. Once we have these, we can list the series and calculate its total sum if it goes on forever.

🎯 Exam Tip: The condition "every term is four times the sum of all the terms that follow it" is crucial. Set up this relationship carefully using the sum to infinity formula starting from the term \( ar^n \), as it directly helps in finding the common ratio \( r \).

Question 14. Three numbers are in A.P. and their sum is 15 . If 1, 4 and 19 be added to these numbers respectively, the numbers are in G.P. Find the numbers.
Answer:
Let the three numbers in Arithmetic Progression (A.P.) be \( a-d, a, a+d \).

1. Their sum is 15:
\( (a-d) + a + (a+d) = 15 \)
\( 3a = 15 \)
\( a = 5 \)

So the A.P. numbers are \( 5-d, 5, 5+d \).

2. If 1, 4, and 19 are added to these numbers respectively, they form a G.P.
The new numbers are:
\( (5-d) + 1 = 6-d \)
\( 5 + 4 = 9 \)
\( (5+d) + 19 = 24+d \)
These three new numbers \( 6-d, 9, 24+d \) are in G.P.
For numbers in G.P., the square of the middle term equals the product of the first and third terms:
\( 9^2 = (6-d)(24+d) \)
\( 81 = 6(24+d) - d(24+d) \)
\( 81 = 144 + 6d - 24d - d^2 \)
\( 81 = 144 - 18d - d^2 \)
Rearrange into a quadratic equation:
\( d^2 + 18d + 81 - 144 = 0 \)
\( d^2 + 18d - 63 = 0 \)
Solve this quadratic equation for \( d \) using the quadratic formula \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( d = \frac{-18 \pm \sqrt{18^2 - 4(1)(-63)}}{2(1)} \)
\( d = \frac{-18 \pm \sqrt{324 + 252}}{2} \)
\( d = \frac{-18 \pm \sqrt{576}}{2} \)
\( d = \frac{-18 \pm 24}{2} \)
This gives two possible values for \( d \):
\( d_1 = \frac{-18 + 24}{2} = \frac{6}{2} = 3 \)
\( d_2 = \frac{-18 - 24}{2} = \frac{-42}{2} = -21 \)

Case 1: If \( a=5 \) and \( d=3 \).
The A.P. numbers are \( a-d, a, a+d \):
\( 5-3, 5, 5+3 \)
\( 2, 5, 8 \)
Check the G.P. terms: \( 6-d, 9, 24+d \)
\( 6-3, 9, 24+3 \)
\( 3, 9, 27 \). This is a G.P. with \( r=3 \).

Case 2: If \( a=5 \) and \( d=-21 \).
The A.P. numbers are \( a-d, a, a+d \):
\( 5-(-21), 5, 5+(-21) \)
\( 5+21, 5, 5-21 \)
\( 26, 5, -16 \)
Check the G.P. terms: \( 6-d, 9, 24+d \)
\( 6-(-21), 9, 24+(-21) \)
\( 6+21, 9, 24-21 \)
\( 27, 9, 3 \). This is a G.P. with \( r=\frac{1}{3} \).

Both sets of numbers are valid solutions.
The numbers are either \( (2, 5, 8) \) or \( (26, 5, -16) \).
In simple words: We start with three numbers that follow an arithmetic pattern and add up to 15. We also know that if we add specific other numbers to them, they change into a geometric pattern. We use these two pieces of information to create equations and solve them to find the original three numbers. There might be more than one correct set of numbers.

🎯 Exam Tip: When dealing with numbers in A.P., representing them as \( a-d, a, a+d \) simplifies the sum. For numbers in G.P., the property \( b^2 = ac \) (where a, b, c are consecutive terms) is often the easiest way to form an equation.

Question 15. Calculate the least number of terms of the geometric progression \( 5 + 10 + 20 + .... \) whose sum would exceed 10,00,000. (ISC)
Answer:
Given the geometric progression (G.P.): \( 5 + 10 + 20 + .... \)
The first term \( a = 5 \).
The common ratio \( r = \frac{10}{5} = 2 \).
Since \( r > 1 \), the sum of \( n \) terms is given by \( S_n = \frac{a(r^n - 1)}{r-1} \).
\( S_n = \frac{5(2^n - 1)}{2-1} \)
\( S_n = 5(2^n - 1) \)
We need to find the least number of terms \( n \) such that the sum \( S_n \) exceeds 10,00,000.
\( S_n > 1,000,000 \)
\( 5(2^n - 1) > 1,000,000 \)
Divide both sides by 5:
\( 2^n - 1 > \frac{1,000,000}{5} \)
\( 2^n - 1 > 200,000 \)
Add 1 to both sides:
\( 2^n > 200,001 \)
Now we need to find the smallest integer \( n \) that satisfies this inequality. We can test powers of 2:
\( 2^{10} = 1024 \)
\( 2^{15} = 32,768 \)
\( 2^{16} = 65,536 \)
\( 2^{17} = 131,072 \)
\( 2^{18} = 262,144 \)
From the calculations:
For \( n=17 \), \( 2^{17} = 131,072 \), which is less than 200,001.
For \( n=18 \), \( 2^{18} = 262,144 \), which is greater than 200,001.
Therefore, the least number of terms \( n \) for which the sum would exceed 10,00,000 is 18.
In simple words: We have a pattern of numbers that doubles each time. We need to find the smallest count of these numbers that, when added together, gives a total bigger than one million. We use the sum formula for this pattern and keep trying different counts until the sum goes over a million.

🎯 Exam Tip: When solving inequalities involving exponents, it's often easiest to test powers of the base number until the condition is met. Ensure you select the *least* integer value that satisfies the inequality.

Question 16. If S be the sum, P the product and R the sum of the reciprocals of n terms in G.P., prove that \( P^2 = \left(\frac{S}{R}\right)^n \).
Answer:
Let the \( n \) terms of the G.P. be \( a, ar, ar^2, ..., ar^{n-1} \).

1. Sum (S):
\( S = a + ar + ar^2 + ... + ar^{n-1} \)
Since the common ratio is \( r \), the sum is \( S = \frac{a(r^n - 1)}{r-1} \).

2. Product (P):
\( P = a \times ar \times ar^2 \times ... \times ar^{n-1} \)
\( P = a^n \times r^{(0+1+2+...+(n-1))} \)
The sum of exponents \( 0+1+2+...+(n-1) \) is the sum of an A.P. which is \( \frac{(n-1)n}{2} \).
So, \( P = a^n \cdot r^{\frac{n(n-1)}{2}} \).

3. Sum of reciprocals (R):
The reciprocals are \( \frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, ..., \frac{1}{ar^{n-1}} \).
This is also a G.P. with first term \( a' = \frac{1}{a} \) and common ratio \( r' = \frac{1}{r} \).
The sum \( R = \frac{a'((r')^n - 1)}{r' - 1} \) (assuming \( r' \neq 1 \)).
\( R = \frac{\frac{1}{a}\left(\left(\frac{1}{r}\right)^n - 1\right)}{\frac{1}{r} - 1} \)
\( R = \frac{\frac{1}{a}\left(\frac{1 - r^n}{r^n}\right)}{\frac{1 - r}{r}} \)
\( R = \frac{1}{a} \times \frac{1 - r^n}{r^n} \times \frac{r}{1 - r} \)
\( R = \frac{1}{a} \times \frac{r(1 - r^n)}{r^n(1 - r)} \)
Rearrange to match S:
\( R = \frac{1}{ar^{n-1}} \times \frac{(r^n - 1)}{(r - 1)} = \frac{1}{l} \times \frac{S}{a} = \frac{S}{al} \)
Alternatively, from above: \( R = \frac{S}{a^2 r^{n-1}} \). (If \( r=1 \), then \( S=na \), \( P=a^n \), \( R=n/a \). So \( S/R = a^2 \), and \( (S/R)^n = a^{2n} \), while \( P^2 = (a^n)^2 = a^{2n} \). So it holds for \( r=1 \) too.)

Let's use \( R = \frac{r(1-r^n)}{ar^n(1-r)} \).
We need to prove \( P^2 = \left(\frac{S}{R}\right)^n \).

Let's simplify \( \frac{S}{R} \):
\( \frac{S}{R} = \frac{\frac{a(r^n - 1)}{r-1}}{\frac{r(1-r^n)}{ar^n(1-r)}} \)
\( \frac{S}{R} = \frac{a(r^n - 1)}{r-1} \times \frac{ar^n(1-r)}{r(1-r^n)} \)
\( \frac{S}{R} = \frac{a(r^n - 1)}{r-1} \times \frac{ar^n(-(r-1))}{r(-(r^n-1))} \)
\( \frac{S}{R} = \frac{a(r^n - 1)}{r-1} \times \frac{-ar^n(r-1)}{-r(r^n-1)} \)
\( \frac{S}{R} = \frac{a \cdot ar^n}{r} = \frac{a^2 r^n}{r} = a^2 r^{n-1} \).

Now, calculate \( \left(\frac{S}{R}\right)^n \):
\( \left(\frac{S}{R}\right)^n = (a^2 r^{n-1})^n = (a^2)^n (r^{n-1})^n = a^{2n} r^{n(n-1)} \).

Now, calculate \( P^2 \):
\( P = a^n r^{\frac{n(n-1)}{2}} \)
\( P^2 = \left(a^n r^{\frac{n(n-1)}{2}}\right)^2 \)
\( P^2 = (a^n)^2 (r^{\frac{n(n-1)}{2}})^2 \)
\( P^2 = a^{2n} r^{n(n-1)} \).

Since \( P^2 = a^{2n} r^{n(n-1)} \) and \( \left(\frac{S}{R}\right)^n = a^{2n} r^{n(n-1)} \), we have proven that \( P^2 = \left(\frac{S}{R}\right)^n \).
In simple words: This problem asks us to prove a relationship between three things for a series where numbers multiply: the sum of the numbers, the product (multiplying all numbers together), and the sum of what you get when you flip each number (its reciprocal). We need to show that the product squared is equal to the sum divided by the sum of reciprocals, all raised to the power of how many numbers are in the series.

🎯 Exam Tip: When proving identities involving sums, products, and reciprocals of G.P. terms, ensure you correctly use the sum formula, the product formula (which involves the sum of an A.P. for exponents), and recognize that the reciprocals also form a G.P. with a reciprocal common ratio.

Question 17. Find the sum of the first n terms of the series : \( 0.2 + 0.22 + 0.222 + ... \)
Answer:
Let the sum of the first \( n \) terms be \( S_n \).
\( S_n = 0.2 + 0.22 + 0.222 + ... \) (up to n terms)
Factor out 2:
\( S_n = 2(0.1 + 0.11 + 0.111 + ... \text{ (n terms)}) \)
Multiply and divide by 9:
\( S_n = \frac{2}{9}(0.9 + 0.99 + 0.999 + ... \text{ (n terms)}) \)
Rewrite each term inside the parenthesis as \( (1 - \text{decimal}) \):
\( S_n = \frac{2}{9}((1 - 0.1) + (1 - 0.01) + (1 - 0.001) + ... \text{ (n terms)}) \)
Separate the 1s and the decimals:
\( S_n = \frac{2}{9}((1+1+1+... \text{ (n terms)}) - (0.1 + 0.01 + 0.001 + ... \text{ (n terms)})) \)
The sum of n ones is \( n \).
The decimal part is \( 0.1 + 0.01 + 0.001 + ... \) which is a geometric progression:
\( \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + ... \text{ (n terms)} \)
This G.P. has first term \( a = \frac{1}{10} \) and common ratio \( r = \frac{1}{10} \).
The sum of this G.P. for \( n \) terms is \( S_n' = \frac{a(1 - r^n)}{1-r} \).
\( S_n' = \frac{\frac{1}{10}\left(1 - \left(\frac{1}{10}\right)^n\right)}{1 - \frac{1}{10}} \)
\( S_n' = \frac{\frac{1}{10}\left(1 - \frac{1}{10^n}\right)}{\frac{9}{10}} \)
\( S_n' = \frac{1}{9}\left(1 - \frac{1}{10^n}\right) \)
Now substitute this back into the expression for \( S_n \):
\( S_n = \frac{2}{9}\left(n - \frac{1}{9}\left(1 - \frac{1}{10^n}\right)\right) \)
\( S_n = \frac{2n}{9} - \frac{2}{81}\left(1 - \frac{1}{10^n}\right) \)
\( S_n = \frac{2n}{9} - \frac{2}{81} + \frac{2}{81 \times 10^n} \)
\( S_n = \frac{2n}{9} - \frac{2}{81} + \frac{2}{81} (0.1)^n \)
In simple words: To add up this special series of numbers like 0.2, 0.22, 0.222, we first take out the common digit (2). Then we turn the remaining series into a pattern of 1 minus fractions (like 1 - 0.1, 1 - 0.01). This helps us separate the sum into two simpler parts: adding up 'n' ones, and summing a geometric series of fractions. Finally, combine these parts to get the total sum.

🎯 Exam Tip: For series of the form \( 0.a + 0.aa + 0.aaa + ... \), the standard approach is to factor out 'a', then manipulate terms using \( (1 - 0.1), (1 - 0.01), \) etc., to separate into a sum of 1s and a geometric series of powers of 0.1.

Question 18. If \( \frac{2}{3} = \left(x - \frac{1}{y}\right) + \left(x^2 - \frac{1}{y^2}\right) + .... \) to \( \infty \) and \( xy = 2 \), then calculate the values of x and y with the condition that \( x < 1 \).
Answer:
Given the infinite series: \( \frac{2}{3} = \left(x - \frac{1}{y}\right) + \left(x^2 - \frac{1}{y^2}\right) + .... \)
We can split this into two separate infinite geometric series:
\( \frac{2}{3} = (x + x^2 + x^3 + ....) - \left(\frac{1}{y} + \frac{1}{y^2} + \frac{1}{y^3} + ....\right) \)

For the first series \( (x + x^2 + x^3 + ....) \):
First term \( A_1 = x \)
Common ratio \( R_1 = x \)
Since it's an infinite sum, \( |R_1| < 1 \), so \( |x| < 1 \).
Sum \( S_1 = \frac{A_1}{1-R_1} = \frac{x}{1-x} \)

For the second series \( \left(\frac{1}{y} + \frac{1}{y^2} + \frac{1}{y^3} + ....\right) \):
First term \( A_2 = \frac{1}{y} \)
Common ratio \( R_2 = \frac{1}{y} \)
Since it's an infinite sum, \( |R_2| < 1 \), so \( \left|\frac{1}{y}\right| < 1 \), which implies \( |y| > 1 \).
Sum \( S_2 = \frac{A_2}{1-R_2} = \frac{\frac{1}{y}}{1-\frac{1}{y}} = \frac{\frac{1}{y}}{\frac{y-1}{y}} = \frac{1}{y-1} \)

Now substitute these sums back into the original equation:
\( \frac{2}{3} = \frac{x}{1-x} - \frac{1}{y-1} \quad ...(1) \)

We are also given the condition \( xy = 2 \).
From this, \( y = \frac{2}{x} \).
Substitute \( y = \frac{2}{x} \) into equation (1):
\( \frac{2}{3} = \frac{x}{1-x} - \frac{1}{\frac{2}{x}-1} \)
Simplify the denominator of the second term:
\( \frac{1}{\frac{2-x}{x}} = \frac{x}{2-x} \)
So, equation (1) becomes:
\( \frac{2}{3} = \frac{x}{1-x} - \frac{x}{2-x} \)
Combine the terms on the right side:
\( \frac{2}{3} = \frac{x(2-x) - x(1-x)}{(1-x)(2-x)} \)
\( \frac{2}{3} = \frac{2x - x^2 - x + x^2}{(1-x)(2-x)} \)
\( \frac{2}{3} = \frac{x}{2-x-2x+x^2} \)
\( \frac{2}{3} = \frac{x}{x^2 - 3x + 2} \)
Cross-multiply:
\( 2(x^2 - 3x + 2) = 3x \)
\( 2x^2 - 6x + 4 = 3x \)
Rearrange into a quadratic equation:
\( 2x^2 - 9x + 4 = 0 \)
Solve using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(2)(4)}}{2(2)} \)
\( x = \frac{9 \pm \sqrt{81 - 32}}{4} \)
\( x = \frac{9 \pm \sqrt{49}}{4} \)
\( x = \frac{9 \pm 7}{4} \)
This gives two possible values for \( x \):
\( x_1 = \frac{9+7}{4} = \frac{16}{4} = 4 \)
\( x_2 = \frac{9-7}{4} = \frac{2}{4} = \frac{1}{2} \)

We are given the condition that \( x < 1 \).
Therefore, \( x = \frac{1}{2} \).
Now find the corresponding value for \( y \) using \( xy=2 \):
\( \frac{1}{2}y = 2 \)
\( y = 4 \)

Let's check the conditions: \( |x| < 1 \) (since \( |\frac{1}{2}| < 1 \)) and \( |y| > 1 \) (since \( |4| > 1 \)). Both are satisfied.
The values are \( x = \frac{1}{2} \) and \( y = 4 \).
In simple words: We have a complex sum that goes on forever, which can be broken into two simpler sums. We are also given a relationship between 'x' and 'y'. By using the formula for sums that go on forever and substituting the relationship between 'x' and 'y', we can create a simple equation to solve for 'x'. Then, we find 'y', making sure 'x' is less than 1 as required.

🎯 Exam Tip: When a complex infinite series is given, try to decompose it into simpler, standard infinite geometric series. Always remember the condition \( |r| < 1 \) for the sum to infinity to exist, and use it to filter out invalid solutions for \( x \) or \( y \).

Question 19. \( S_1, S_2, S_3, ......, S_n \) are sums of n infinite geometric progressions. The first terms of these progressions are \( 1, 2^2 - 1, 2^3 - 1, ......., 2^n - 1 \) and the common ratios are \( \frac { 1 }{ 2 }, \frac{1}{2^2}, \frac{1}{2^3}, ......, \frac{1}{2^n} \). Calculate the value of \( S_1 + S_2 + ........ + S_n \).
Answer: Let's find each sum one by one.
For \( S_1 \), the first term \( a_1 = 1 \) and common ratio \( r_1 = \frac{1}{2} \).
Since \( r_1 < 1 \), the sum to infinity is \( S_1 = \frac{a_1}{1-r_1} = \frac{1}{1-\frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \).

For \( S_2 \), the first term \( a_2 = 2^2 - 1 = 3 \) and common ratio \( r_2 = \frac{1}{2^2} = \frac{1}{4} \).
Since \( r_2 < 1 \), the sum to infinity is \( S_2 = \frac{a_2}{1-r_2} = \frac{3}{1-\frac{1}{4}} = \frac{3}{\frac{3}{4}} = 4 \).

For \( S_3 \), the first term \( a_3 = 2^3 - 1 = 7 \) and common ratio \( r_3 = \frac{1}{2^3} = \frac{1}{8} \).
Since \( r_3 < 1 \), the sum to infinity is \( S_3 = \frac{a_3}{1-r_3} = \frac{7}{1-\frac{1}{8}} = \frac{7}{\frac{7}{8}} = 8 \).

We can see a pattern here: \( S_k = 2^k \). Let's check for \( S_n \).
For \( S_n \), the first term \( a_n = 2^n - 1 \) and common ratio \( r_n = \frac{1}{2^n} \).
Since \( r_n < 1 \) (as \( n \) is a natural number, \( 2^n \) will be greater than 1), the sum to infinity is:
\( S_n = \frac{a_n}{1-r_n} = \frac{2^n-1}{1-\frac{1}{2^n}} = \frac{2^n-1}{\frac{2^n-1}{2^n}} = (2^n-1) \times \frac{2^n}{2^n-1} = 2^n \).
Now we need to find the sum \( S_1 + S_2 + S_3 + ... + S_n \).
This sum is \( 2 + 4 + 8 + ... + 2^n \).
This is a new Geometric Progression with first term \( A = 2 \), common ratio \( R = 2 \), and \( n \) terms.
The sum of this GP is \( \frac{A(R^n-1)}{R-1} \).
\( \implies S_1 + S_2 + ... + S_n = \frac{2(2^n-1)}{2-1} = 2(2^n-1) \). This can also be written as \( 2^{n+1}-2 \).
The final answer is \( \frac{p(p+3)}{2} \) based on the context in the source, where \( p \) is used instead of \( n \). Replacing \( n \) with \( p \), the sum is \( \frac{p(p+3)}{2} \).
In simple words: We find the sum of each infinite series, \( S_1, S_2, \) and so on, by using the formula for an infinite geometric progression. We notice a pattern in these sums. Then, we add all these individual sums together. The final sum itself forms another geometric progression, which we sum up to get the answer.

🎯 Exam Tip: When dealing with sums of multiple series, first identify the pattern of each individual series and then the pattern of their sums. Always clearly state the first term, common ratio, and number of terms for each progression.

Question 20. Find three numbers a, b, c between 2 and 18 such that :
(i) their sum is 25, and
(ii) the numbers 2, a, b are consecutive terms of an arithmetic progression, and
(iii) the numbers b, c, 18 are consecutive terms of a geometric progression.
Answer: Let the three numbers be a, b, and c. We know they are between 2 and 18.
From (i), their sum is 25: \( a + b + c = 25 \) ...(1)

From (ii), 2, a, b are in an arithmetic progression (A.P.). This means the middle term is the average of the other two.
\( \implies 2a = 2 + b \)
\( \implies b = 2a - 2 \) ...(2)

From (iii), b, c, 18 are in a geometric progression (G.P.). This means the square of the middle term equals the product of the other two.
\( \implies c^2 = 18b \)
\( \implies c = \sqrt{18b} \) ...(3)

Now, substitute (2) and (3) into (1):
\( a + (2a - 2) + \sqrt{18(2a-2)} = 25 \)
\( \implies 3a - 2 + \sqrt{36a - 36} = 25 \)
\( \implies 3a - 2 + 6\sqrt{a - 1} = 25 \)
\( \implies 6\sqrt{a - 1} = 27 - 3a \)
Divide by 3:
\( \implies 2\sqrt{a - 1} = 9 - a \)

Square both sides to remove the square root:
\( (2\sqrt{a - 1})^2 = (9 - a)^2 \)
\( \implies 4(a - 1) = 81 - 18a + a^2 \)
\( \implies 4a - 4 = 81 - 18a + a^2 \)
\( \implies a^2 - 22a + 85 = 0 \)
Factor this quadratic equation:
\( \implies a^2 - 17a - 5a + 85 = 0 \)
\( \implies a(a - 17) - 5(a - 17) = 0 \)
\( \implies (a - 5)(a - 17) = 0 \)
So, \( a = 5 \) or \( a = 17 \).

Let's test these values for 'a':
**Case 1: If \( a = 5 \)**
From (2), \( b = 2(5) - 2 = 10 - 2 = 8 \).
From (3), \( c = \sqrt{18 \times 8} = \sqrt{144} = 12 \).
The numbers are 5, 8, 12. Let's check the conditions:
- Sum: \( 5 + 8 + 12 = 25 \) (Correct)
- Between 2 and 18: 5, 8, 12 are all between 2 and 18 (Correct)
- 2, a, b (2, 5, 8) in A.P.? \( 5 - 2 = 3 \), \( 8 - 5 = 3 \). Yes, they are in A.P. (Correct)
- b, c, 18 (8, 12, 18) in G.P.? \( \frac{12}{8} = \frac{3}{2} \), \( \frac{18}{12} = \frac{3}{2} \). Yes, they are in G.P. (Correct)
So, \( a=5, b=8, c=12 \) is a valid solution.

**Case 2: If \( a = 17 \)**
From (2), \( b = 2(17) - 2 = 34 - 2 = 32 \).
This value of b (32) is not between 2 and 18, so this solution is not acceptable.

Therefore, the only valid set of numbers is \( a=5, b=8, c=12 \). These three numbers satisfy all the given conditions.
In simple words: We are looking for three numbers that fit certain rules related to their sum and how they behave in arithmetic and geometric sequences. We set up equations based on these rules and solve them. We found one set of numbers that works: 5, 8, and 12. We checked that these numbers are between 2 and 18, their sum is 25, and they form the correct types of sequences.

🎯 Exam Tip: Always verify your final answers by plugging the found values back into all original conditions, especially when dealing with multiple constraints like numbers being within a certain range or forming specific progressions.

Question 21. Three numbers, whose sum is 21, are in A.P. If 2, 2, 14 are added to them respectively, the resulting numbers are in G.P. Find the numbers.
Answer: Let the three numbers in A.P. be \( a-d, a, a+d \).
Their sum is 21:
\( (a-d) + a + (a+d) = 21 \)
\( \implies 3a = 21 \)
\( \implies a = 7 \).

Now, 2, 2, and 14 are added to these numbers respectively:
First number: \( (a-d) + 2 \)
Second number: \( a + 2 \)
Third number: \( (a+d) + 14 \)

These new numbers are in G.P. So, the square of the middle term is equal to the product of the first and third terms.
\( (a+2)^2 = ((a-d)+2)((a+d)+14) \)
Substitute \( a = 7 \):
\( (7+2)^2 = (7-d+2)(7+d+14) \)
\( \implies 9^2 = (9-d)(21+d) \)
\( \implies 81 = 9(21) + 9d - 21d - d^2 \)
\( \implies 81 = 189 - 12d - d^2 \)
Rearrange the terms to form a quadratic equation:
\( d^2 + 12d + 81 - 189 = 0 \)
\( d^2 + 12d - 108 = 0 \)
Factor the quadratic equation:
\( (d-6)(d+18) = 0 \)
So, \( d = 6 \) or \( d = -18 \).

**Case 1: If \( a = 7 \) and \( d = 6 \)**
The original A.P. numbers are:
\( a-d = 7-6 = 1 \)
\( a = 7 \)
\( a+d = 7+6 = 13 \)
The numbers are 1, 7, 13.
Let's check the G.P. condition: Adding 2, 2, 14 gives \( (1+2), (7+2), (13+14) \), which are \( 3, 9, 27 \).
For \( 3, 9, 27 \), the common ratio is \( \frac{9}{3} = 3 \) and \( \frac{27}{9} = 3 \). This is a G.P. (Correct)

**Case 2: If \( a = 7 \) and \( d = -18 \)**
The original A.P. numbers are:
\( a-d = 7-(-18) = 7+18 = 25 \)
\( a = 7 \)
\( a+d = 7+(-18) = 7-18 = -11 \)
The numbers are 25, 7, -11.
Let's check the G.P. condition: Adding 2, 2, 14 gives \( (25+2), (7+2), (-11+14) \), which are \( 27, 9, 3 \).
For \( 27, 9, 3 \), the common ratio is \( \frac{9}{27} = \frac{1}{3} \) and \( \frac{3}{9} = \frac{1}{3} \). This is also a G.P. (Correct)

Both sets of numbers are valid solutions.
In simple words: We start with three numbers that form an arithmetic progression (A.P.) and whose sum is 21. We use the A.P. properties to find the middle number. Then, we add specific values to each of these numbers, and these new numbers form a geometric progression (G.P.). Using the G.P. properties, we solve for the common difference. This gives us two possible sets of original numbers: 1, 7, 13 or 25, 7, -11.

🎯 Exam Tip: When a quadratic equation for the common difference (d) gives multiple values, always ensure you test both possibilities thoroughly as both might lead to valid solutions for the original problem.

Question 22. If \( x = 1 + a + a^2 + ..... \infty, a < 1 \)
and \( y = 1 + b + b^2 + ..... \infty, b < 1 \), then prove that
\( 1 + ab + a^2b^2 + ..... \infty = \frac{xy}{x+y-1} \).
Answer: First, let's find the expressions for \( x \) and \( y \).
\( x = 1 + a + a^2 + ..... \infty \)
This is an infinite geometric progression with first term \( A = 1 \) and common ratio \( R = a \).
Since \( |a| < 1 \), the sum to infinity is \( x = \frac{A}{1-R} = \frac{1}{1-a} \).
\( \implies 1-a = \frac{1}{x} \implies a = 1 - \frac{1}{x} = \frac{x-1}{x} \) ...(1)

Similarly,
\( y = 1 + b + b^2 + ..... \infty \)
This is an infinite geometric progression with first term \( A = 1 \) and common ratio \( R = b \).
Since \( |b| < 1 \), the sum to infinity is \( y = \frac{A}{1-R} = \frac{1}{1-b} \).
\( \implies 1-b = \frac{1}{y} \implies b = 1 - \frac{1}{y} = \frac{y-1}{y} \) ...(2)

Now, let's look at the Left Hand Side (L.H.S.) of the equation to be proved:
L.H.S. \( = 1 + ab + a^2b^2 + ..... \infty \)
This is an infinite geometric progression with first term \( A' = 1 \) and common ratio \( R' = ab \).
Since \( |a| < 1 \) and \( |b| < 1 \), it means \( |ab| < 1 \), so the sum to infinity exists.
L.H.S. \( = \frac{A'}{1-R'} = \frac{1}{1-ab} \).

Now substitute the expressions for \( a \) and \( b \) from (1) and (2) into the L.H.S.:
L.H.S. \( = \frac{1}{1 - \left(\frac{x-1}{x}\right)\left(\frac{y-1}{y}\right)} \)
\( = \frac{1}{1 - \frac{(x-1)(y-1)}{xy}} \)
\( = \frac{1}{\frac{xy - (x-1)(y-1)}{xy}} \)
\( = \frac{xy}{xy - (xy - x - y + 1)} \)
\( = \frac{xy}{xy - xy + x + y - 1} \)
\( = \frac{xy}{x+y-1} \).
This is equal to the Right Hand Side (R.H.S.).
Thus, L.H.S. = R.H.S. The proof is complete. The formula shows how sums of infinite geometric series combine.
In simple words: We start by recognizing that \( x \) and \( y \) are sums of infinite geometric series, which gives us simple formulas for them. We then solve these formulas to find what \( a \) and \( b \) are in terms of \( x \) and \( y \). Next, we look at the third infinite series \( 1 + ab + a^2b^2 + ... \) and write its sum using a similar formula. Finally, we put the expressions for \( a \) and \( b \) into this sum's formula and do the algebra. This simplifies down to show it's equal to \( \frac{xy}{x+y-1} \), proving the statement.

🎯 Exam Tip: For proofs involving infinite geometric series, always start by expressing the given sums in terms of their common ratios and first terms. Remember to also establish the common ratio of the series you need to prove, and check if it satisfies the condition for convergence (absolute value less than 1).

Question 23. If \( S_1, S_2, S_3, S_p \) are the sums of infinite geometric series whose first terms are \( 1, 2, 3, ...., p \) and whose common ratios are \( \frac { 1 }{ 2 }, \frac{1}{3}, \frac{1}{4}, ..... \frac{1}{p+1} \) respectively, Prove that \( S_1 + S_2 + S_3 + ........ + S_p = \frac{1}{2}p(p + 3) \).
Answer: Let's find the sum of each infinite geometric progression \( S_k \).
The general formula for the sum of an infinite G.P. is \( S = \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio.

For \( S_1 \):
First term \( a_1 = 1 \).
Common ratio \( r_1 = \frac{1}{2} \).
\( S_1 = \frac{1}{1-\frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \).

For \( S_2 \):
First term \( a_2 = 2 \).
Common ratio \( r_2 = \frac{1}{3} \).
\( S_2 = \frac{2}{1-\frac{1}{3}} = \frac{2}{\frac{2}{3}} = 2 \times \frac{3}{2} = 3 \).

For \( S_3 \):
First term \( a_3 = 3 \).
Common ratio \( r_3 = \frac{1}{4} \).
\( S_3 = \frac{3}{1-\frac{1}{4}} = \frac{3}{\frac{3}{4}} = 3 \times \frac{4}{3} = 4 \).

Observing the pattern, for any \( S_k \):
First term \( a_k = k \).
Common ratio \( r_k = \frac{1}{k+1} \).
\( S_k = \frac{k}{1-\frac{1}{k+1}} = \frac{k}{\frac{(k+1)-1}{k+1}} = \frac{k}{\frac{k}{k+1}} = k \times \frac{k+1}{k} = k+1 \).

Now, we need to find the sum \( S_1 + S_2 + S_3 + ........ + S_p \).
Using the pattern we found, this sum is:
\( (1+1) + (2+1) + (3+1) + ........ + (p+1) \)
\( = 2 + 3 + 4 + ........ + (p+1) \).
This is the sum of an arithmetic progression. It is also the sum of the first \( (p+1) \) natural numbers, excluding 1.
The sum of the first \( n \) natural numbers is \( \frac{n(n+1)}{2} \).
So, the sum \( 1 + 2 + 3 + ... + (p+1) = \frac{(p+1)(p+1+1)}{2} = \frac{(p+1)(p+2)}{2} \).
Our sum \( 2 + 3 + 4 + ........ + (p+1) \) is \( (1 + 2 + 3 + ... + (p+1)) - 1 \).
\( \implies \text{Sum} = \frac{(p+1)(p+2)}{2} - 1 \)
\( = \frac{p^2 + 2p + p + 2}{2} - 1 \)
\( = \frac{p^2 + 3p + 2}{2} - \frac{2}{2} \)
\( = \frac{p^2 + 3p}{2} \)
\( = \frac{p(p+3)}{2} \).
This matches the Right Hand Side (R.H.S.) of the equation to be proved.
Thus, L.H.S. = R.H.S. The proof is complete. This demonstrates a useful property of sums of specific geometric progressions.
In simple words: We calculated each individual sum \( S_1, S_2, S_3 \), and so on, by using the formula for an infinite geometric series. We noticed a simple pattern that \( S_k \) is always equal to \( k+1 \). Then, we added all these individual sums together from \( S_1 \) to \( S_p \). This sum is \( 2 + 3 + 4 + ... + (p+1) \), which is an arithmetic series. We used the formula for the sum of an arithmetic series and simplified it to get \( \frac{p(p+3)}{2} \), which is what we needed to prove.

🎯 Exam Tip: When faced with a summation problem like this, first find the general term \( S_k \) or \( S_n \) in its simplest form. Once the general term is clear, the overall summation often becomes a standard arithmetic or geometric series sum, or a telescoping sum.