OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Chapter Test

S Chand Class 11 ICSE Maths Solutions Chapter 14 Sequence and Series Chapter Test

 

Question 1. Write down the first five terms of the sequence, whose nth term is \( (-1)^{n-1} \cdot 5^{n+1} \)
Answer: The \( n^{th} \) term of the sequence is given by \( T_n = (-1)^{n-1} \cdot 5^{n+1} \). We can find the first five terms by substituting \( n=1, 2, 3, 4, 5 \).
For \( n=1 \): \( T_1 = (-1)^{1-1} \cdot 5^{1+1} = (-1)^0 \cdot 5^2 = 1 \cdot 25 = 25 \)
For \( n=2 \): \( T_2 = (-1)^{2-1} \cdot 5^{2+1} = (-1)^1 \cdot 5^3 = -1 \cdot 125 = -125 \)
For \( n=3 \): \( T_3 = (-1)^{3-1} \cdot 5^{3+1} = (-1)^2 \cdot 5^4 = 1 \cdot 625 = 625 \)
For \( n=4 \): \( T_4 = (-1)^{4-1} \cdot 5^{4+1} = (-1)^3 \cdot 5^5 = -1 \cdot 3125 = -3125 \)
For \( n=5 \): \( T_5 = (-1)^{5-1} \cdot 5^{5+1} = (-1)^4 \cdot 5^6 = 1 \cdot 15625 = 15625 \)
So, the first five terms are 25, -125, 625, -3125, 15625. Notice the alternating sign in each term.
In simple words: To find each term, put the term number (like 1 for the first, 2 for the second, etc.) into the given formula for 'n'. Calculate the power of -1 and the power of 5, then multiply them to get the term's value.

🎯 Exam Tip: When evaluating terms with \( (-1)^{n-1} \), remember that \( (-1)^{\text{even number}} = 1 \) and \( (-1)^{\text{odd number}} = -1 \). Be careful with exponent calculations.

 

Question 2. If the 3rd and the 6th terms of an A.P. and 7 and 13 respectively, find the first term and the common difference.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the arithmetic progression (A.P.).
The formula for the \( n^{th} \) term of an A.P. is \( T_n = a + (n-1)d \).
We are given that the 3rd term is 7:
\( T_3 = a + (3-1)d = a + 2d = 7 \)...(1)
We are also given that the 6th term is 13:
\( T_6 = a + (6-1)d = a + 5d = 13 \)...(2)
Now, subtract equation (1) from equation (2) to eliminate \( a \):
\( (a + 5d) - (a + 2d) = 13 - 7 \)
\( 3d = 6 \)
\( d = \frac{6}{3} \)
\( d = 2 \)
Substitute the value of \( d=2 \) into equation (1):
\( a + 2(2) = 7 \)
\( a + 4 = 7 \)
\( a = 7 - 4 \)
\( a = 3 \)
Thus, the first term of the A.P. is 3 and the common difference is 2. The arithmetic progression has a constant difference between consecutive terms.
In simple words: We used the given 3rd and 6th terms to create two simple math problems with 'a' (first term) and 'd' (common difference). By solving these two problems together, we found that the first term is 3 and the common difference is 2.

🎯 Exam Tip: Always set up simultaneous equations correctly from the given terms. Subtracting the equations is a common way to quickly find the common difference.

 

Question 3. Find the sum of all natural numbers between 100 and 1000 which are multiple of 5.
Answer: We need to find the natural numbers between 100 and 1000 that are multiples of 5. "Between" means we do not include 100 or 1000.
The first multiple of 5 greater than 100 is 105.
The last multiple of 5 less than 1000 is 995.
So, the sequence of numbers is 105, 110, 115, ..., 995.
This is an arithmetic progression (A.P.) with:
First term \( a = 105 \)
Common difference \( d = 110 - 105 = 5 \)
Last term \( l = T_n = 995 \)
First, we find the number of terms \( n \) using the formula \( T_n = a + (n-1)d \):
\( 995 = 105 + (n-1)5 \)
\( 995 - 105 = (n-1)5 \)
\( 890 = (n-1)5 \)
\( n-1 = \frac{890}{5} \)
\( n-1 = 178 \)
\( n = 178 + 1 \)
\( n = 179 \)
Next, we find the sum of these \( n \) terms using the formula \( S_n = \frac{n}{2}(a+l) \):
\( S_{179} = \frac{179}{2}(105 + 995) \)
\( S_{179} = \frac{179}{2}(1100) \)
\( S_{179} = 179 \times 550 \)
\( S_{179} = 98450 \)
Therefore, the sum of all natural numbers between 100 and 1000 which are multiples of 5 is 98450.
In simple words: First, we listed the numbers that fit the rule. These numbers form a pattern called an arithmetic progression. We found how many numbers are in this list and then used a special formula to quickly add all of them together.

🎯 Exam Tip: For problems with "between" a range, carefully identify the first and last terms to be included. Ensure all variables (a, d, n, l) are correctly determined before calculating the sum.

 

Question 4. How many terms of the A.P., -6, \( \frac { -11 }{ 2 } \), -5, ...... are needed to give the sum -25 ?
Answer: Given the arithmetic progression (A.P.) is \( -6, -\frac{11}{2}, -5, \ldots \).
The first term is \( a = -6 \).
The common difference \( d \) is the second term minus the first term:
\( d = -\frac{11}{2} - (-6) = -\frac{11}{2} + 6 = \frac{-11 + 12}{2} = \frac{1}{2} \)
We are given that the sum of \( n \) terms is \( S_n = -25 \).
The formula for the sum of \( n \) terms of an A.P. is \( S_n = \frac{n}{2}[2a + (n-1)d] \).
Substitute the known values into the formula:
\( -25 = \frac{n}{2}[2(-6) + (n-1)\frac{1}{2}] \)
Multiply both sides by 2:
\( -50 = n[-12 + \frac{n-1}{2}] \)
Multiply the term inside the bracket by 2 to clear the fraction:
\( -100 = n[-24 + (n-1)] \)
\( -100 = n[-24 + n - 1] \)
\( -100 = n(n - 25) \)
\( -100 = n^2 - 25n \)
Rearrange into a quadratic equation:
\( n^2 - 25n + 100 = 0 \)
Factorize the quadratic equation:
We need two numbers that multiply to 100 and add up to -25. These are -5 and -20.
\( (n-5)(n-20) = 0 \)
This gives two possible values for \( n \):
\( n-5 = 0 \implies n = 5 \)
\( n-20 = 0 \implies n = 20 \)
Both values are positive integers, so both are valid. This sequence has both positive and negative terms, and the sum can reach -25 in two different ways. Sometimes there can be two valid answers for the number of terms.
In simple words: We figured out the starting number and the pattern (common difference). Then, we put these into the sum formula and solved it like a puzzle. We found that either 5 terms or 20 terms add up to -25.

🎯 Exam Tip: Be very careful with arithmetic, especially signs, when finding the common difference. When solving for 'n' in the sum formula, you might get a quadratic equation with two valid positive integer solutions.

 

Question 5. Determine the sum of the first 35 terms of an A.P. if a2 = 2 and a7 = 22.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
We know that the \( n^{th} \) term of an A.P. is \( a_n = a + (n-1)d \).
Given that the 2nd term \( a_2 = 2 \):
\( a + (2-1)d = 2 \implies a + d = 2 \)...(1)
Given that the 7th term \( a_7 = 22 \):
\( a + (7-1)d = 22 \implies a + 6d = 22 \)...(2)
To find \( d \), subtract equation (1) from equation (2):
\( (a + 6d) - (a + d) = 22 - 2 \)
\( 5d = 20 \)
\( d = \frac{20}{5} \)
\( d = 4 \)
Now, substitute the value of \( d=4 \) into equation (1) to find \( a \):
\( a + 4 = 2 \)
\( a = 2 - 4 \)
\( a = -2 \)
We need to find the sum of the first 35 terms, \( S_{35} \). The formula for the sum of \( n \) terms is \( S_n = \frac{n}{2}[2a + (n-1)d] \).
Substitute \( n=35 \), \( a=-2 \), and \( d=4 \):
\( S_{35} = \frac{35}{2}[2(-2) + (35-1)4] \)
\( S_{35} = \frac{35}{2}[-4 + (34)4] \)
\( S_{35} = \frac{35}{2}[-4 + 136] \)
\( S_{35} = \frac{35}{2}[132] \)
\( S_{35} = 35 \times 66 \)
\( S_{35} = 2310 \)
The sum of an arithmetic progression depends on the first term, common difference, and number of terms.
In simple words: First, we used the given 2nd and 7th terms to find the starting number (first term) and the pattern (common difference) of the series. Once we knew these, we used a special formula to quickly add up the first 35 terms.

🎯 Exam Tip: Ensure accurate calculations for the first term and common difference, as any error in these will propagate through the sum calculation. Organize your steps clearly.

 

Question 6. If the first term of an A.P. is 2 and the sum of first five terms is equal to one-fourth of the sum of the next five terms, show that the 200th term is -112.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
We are given \( a = 2 \).
The sum of the first five terms is \( S_5 \).
The sum of the next five terms (from the 6th term to the 10th term) is \( S_{10} - S_5 \).
According to the problem, \( S_5 = \frac{1}{4}(S_{10} - S_5) \).
Multiply both sides by 4: \( 4S_5 = S_{10} - S_5 \)
Rearrange the equation: \( 5S_5 = S_{10} \)
Now, use the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
For \( S_5 \): \( S_5 = \frac{5}{2}[2(2) + (5-1)d] = \frac{5}{2}[4 + 4d] = 5(2 + 2d) = 10 + 10d \).
For \( S_{10} \): \( S_{10} = \frac{10}{2}[2(2) + (10-1)d] = 5[4 + 9d] = 20 + 45d \).
Substitute these into \( 5S_5 = S_{10} \):
\( 5(10 + 10d) = 20 + 45d \)
\( 50 + 50d = 20 + 45d \)
\( 50d - 45d = 20 - 50 \)
\( 5d = -30 \)
\( d = \frac{-30}{5} \)
\( d = -6 \)
Now, we need to find the 200th term, \( T_{200} \). The formula for the \( n^{th} \) term is \( T_n = a + (n-1)d \).
\( T_{200} = a + (200-1)d = 2 + 199(-6) \)
\( T_{200} = 2 - 1194 \)
\( T_{200} = -1192 \)
The sum formula is very useful as it connects the first term and common difference to the total sum over 'n' terms.
In simple words: We used the given facts about the first term and the relation between the sums of terms to find the common difference of the pattern. Once we found the pattern, we calculated the 200th number in the sequence. Based on our calculation, the 200th term is -1192.

🎯 Exam Tip: Be careful when dealing with sums of 'next' terms; it's often easier to express them as a difference of total sums (e.g., sum of next five terms is \( S_{10} - S_5 \)). Ensure your final calculation for the \( n^{th} \) term is correct.

 

Question 7. Insert 3 arithmetic means between 2 and 10.
Answer: To insert 3 arithmetic means between 2 and 10, we form an A.P. like this:
\( 2, A_1, A_2, A_3, 10 \)
In this sequence, the first term is \( a = 2 \) and the last term is \( T_5 = 10 \). The total number of terms is \( n = 5 \).
Using the formula for the \( n^{th} \) term of an A.P., \( T_n = a + (n-1)d \):
\( T_5 = a + (5-1)d \)
\( 10 = 2 + 4d \)
\( 10 - 2 = 4d \)
\( 8 = 4d \)
\( d = \frac{8}{4} \)
\( d = 2 \)
Now that we have the common difference \( d=2 \), we can find the arithmetic means:
\( A_1 = T_2 = a + d = 2 + 2 = 4 \)
\( A_2 = T_3 = a + 2d = 2 + 2(2) = 2 + 4 = 6 \)
\( A_3 = T_4 = a + 3d = 2 + 3(2) = 2 + 6 = 8 \)
Thus, the three arithmetic means between 2 and 10 are 4, 6, and 8. Arithmetic means are terms that fit perfectly into an arithmetic sequence between two given numbers.
In simple words: We need to find three numbers that create an even pattern between 2 and 10. We found that each number should go up by 2. So, starting from 2, we add 2 three times: 2+2=4, 4+2=6, 6+2=8.

🎯 Exam Tip: Remember that if you insert 'k' arithmetic means between two numbers, the total number of terms in the new sequence will be \( k+2 \).

 

Question 8. Find 12 th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Answer: Let \( a \) be the first term and \( r \) be the common ratio of the geometric progression (G.P.).
The formula for the \( n^{th} \) term of a G.P. is \( T_n = ar^{n-1} \).
We are given the common ratio \( r = 2 \).
We are given that the 8th term \( T_8 = 192 \):
\( T_8 = ar^{8-1} = ar^7 = 192 \)
Substitute \( r=2 \) into the equation:
\( a(2^7) = 192 \)
\( a(128) = 192 \)
\( a = \frac{192}{128} \)
Divide both numerator and denominator by their greatest common divisor (64):
\( a = \frac{3}{2} \)
Now we need to find the 12th term, \( T_{12} \):
\( T_{12} = ar^{12-1} = ar^{11} \)
Substitute \( a=\frac{3}{2} \) and \( r=2 \):
\( T_{12} = \frac{3}{2} \cdot 2^{11} \)
\( T_{12} = 3 \cdot 2^{11-1} \)
\( T_{12} = 3 \cdot 2^{10} \)
\( T_{12} = 3 \cdot 1024 \)
\( T_{12} = 3072 \)
Geometric progressions (G.P.) involve multiplying by a constant ratio to get the next term.
In simple words: We used the 8th term and the common ratio to find the very first number in the sequence. Once we had that, we used the common ratio again to jump ahead and find the 12th number.

🎯 Exam Tip: Remember the general formula for the \( n^{th} \) term of a G.P. (\( ar^{n-1} \)). Be careful with exponent calculations when finding higher terms.

 

Question 9. The first term of a G.P. is 1. The sum of the third and fifth terms is 90. Find the common ratio of the G.P.
Answer: Let \( a \) be the first term and \( r \) be the common ratio of the G.P.
We are given that the first term \( a = 1 \).
The formula for the \( n^{th} \) term of a G.P. is \( T_n = ar^{n-1} \).
The third term is \( T_3 = ar^{3-1} = ar^2 \).
The fifth term is \( T_5 = ar^{5-1} = ar^4 \).
We are given that the sum of the third and fifth terms is 90:
\( T_3 + T_5 = 90 \)
\( ar^2 + ar^4 = 90 \)
Substitute \( a=1 \) into the equation:
\( 1 \cdot r^2 + 1 \cdot r^4 = 90 \)
\( r^4 + r^2 = 90 \)
Rearrange this into a quadratic form by letting \( x = r^2 \):
\( x^2 + x - 90 = 0 \)
Factorize the quadratic equation:
We need two numbers that multiply to -90 and add up to 1. These numbers are 10 and -9.
\( (x + 10)(x - 9) = 0 \)
This gives two possible values for \( x \):
\( x + 10 = 0 \implies x = -10 \)
\( x - 9 = 0 \implies x = 9 \)
Since \( x = r^2 \), and \( r \) must be a real number, \( r^2 \) cannot be negative.
Therefore, \( r^2 = -10 \) has no real solution.
So, we take \( r^2 = 9 \).
Taking the square root of both sides:
\( r = \pm \sqrt{9} \)
\( r = \pm 3 \)
The common ratio can be positive or negative, which affects whether the terms alternate in sign or not.
In simple words: We put the first term and the formula for terms into the given sum. This gave us a puzzle to solve for 'r' squared. We found two possible values for 'r' squared, but only one gave real numbers for 'r'. So, the common ratio can be either 3 or -3.

🎯 Exam Tip: When solving for the common ratio \( r \) from \( r^2 \), remember to consider both positive and negative roots (\( \pm \)). Also, ensure the common ratio results in real terms if no complex numbers are specified.

 

Question 10. The sum of first three terms of a G.P. is \( \frac { 39 }{ 10 } \) and their product is 1 . Find the common ratio and the terms.
Answer: Let the first three terms of the G.P. be \( \frac{a}{r}, a, ar \). This way simplifies calculations when the product is given.
The product of these three terms is 1:
\( \frac{a}{r} \cdot a \cdot ar = 1 \)
\( a^3 = 1 \)
Taking the cube root of both sides:
\( a = 1 \)
The sum of the first three terms is \( \frac{39}{10} \):
\( \frac{a}{r} + a + ar = \frac{39}{10} \)
Substitute \( a=1 \) into the sum equation:
\( \frac{1}{r} + 1 + r = \frac{39}{10} \)
Multiply the entire equation by \( 10r \) to clear the denominators (assuming \( r \neq 0 \)):
\( 10(1) + 10r + 10r^2 = 39r \)
\( 10 + 10r + 10r^2 = 39r \)
Rearrange into a quadratic equation:
\( 10r^2 + 10r - 39r + 10 = 0 \)
\( 10r^2 - 29r + 10 = 0 \)
Factorize the quadratic equation. We need two numbers that multiply to \( 10 \times 10 = 100 \) and add up to -29. These are -25 and -4.
\( 10r^2 - 25r - 4r + 10 = 0 \)
Factor by grouping:
\( 5r(2r - 5) - 2(2r - 5) = 0 \)
\( (5r - 2)(2r - 5) = 0 \)
This gives two possible values for \( r \):
\( 5r - 2 = 0 \implies 5r = 2 \implies r = \frac{2}{5} \)
\( 2r - 5 = 0 \implies 2r = 5 \implies r = \frac{5}{2} \)
Now, find the terms for each value of \( r \) when \( a=1 \):
Case 1: If \( r = \frac{2}{5} \)
The terms are \( \frac{a}{r}, a, ar \):
\( \frac{1}{2/5}, 1, 1 \cdot \frac{2}{5} \)
\( \frac{5}{2}, 1, \frac{2}{5} \)
Case 2: If \( r = \frac{5}{2} \)
The terms are \( \frac{a}{r}, a, ar \):
\( \frac{1}{5/2}, 1, 1 \cdot \frac{5}{2} \)
\( \frac{2}{5}, 1, \frac{5}{2} \)
This clever way of writing the terms (a/r, a, ar) is often helpful when the product of terms in a G.P. is given.
In simple words: We used a trick to represent the three terms so that their product directly gave us the first term, 'a'. Then, using the sum of these terms, we found two possible common ratios, 'r'. Finally, we listed the three terms for each possible 'r' value.

🎯 Exam Tip: When the product of terms in a G.P. is given, representing the terms as \( \frac{a}{r}, a, ar \) (for 3 terms) or \( \frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2 \) (for 5 terms) simplifies finding the first term immediately.

 

Question 11. The sum of some terms of a G.P. is 315 and the first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.
Answer: Let \( a \) be the first term, \( r \) be the common ratio, \( n \) be the number of terms, and \( S_n \) be the sum of \( n \) terms of the G.P.
We are given:
\( S_n = 315 \)
\( a = 5 \)
\( r = 2 \)
First, we find the number of terms \( n \) using the formula for the sum of \( n \) terms of a G.P. (since \( r > 1 \)):
\( S_n = \frac{a(r^n - 1)}{r-1} \)
Substitute the given values:
\( 315 = \frac{5(2^n - 1)}{2-1} \)
\( 315 = 5(2^n - 1) \)
Divide both sides by 5:
\( \frac{315}{5} = 2^n - 1 \)
\( 63 = 2^n - 1 \)
Add 1 to both sides:
\( 64 = 2^n \)
We know that \( 2^6 = 64 \), so:
\( 2^6 = 2^n \implies n = 6 \)
So, there are 6 terms in the G.P.
Next, we find the last term, which is the \( n^{th} \) term, \( T_n \). The formula for the \( n^{th} \) term of a G.P. is \( T_n = ar^{n-1} \).
Substitute \( n=6 \), \( a=5 \), and \( r=2 \):
\( T_6 = 5 \cdot 2^{6-1} \)
\( T_6 = 5 \cdot 2^5 \)
\( T_6 = 5 \cdot 32 \)
\( T_6 = 160 \)
The sum of a finite geometric progression increases rapidly, especially when the common ratio is greater than 1.
In simple words: We used the given sum, first term, and common ratio in the G.P. sum formula to find how many terms are in the series. After that, we used the G.P. term formula to calculate the value of the very last term.

🎯 Exam Tip: When using the sum formula to find 'n', remember to solve for the exponent. It often involves recognizing powers of the common ratio.

 

Question 12. Find the sum of the series 0.6 + 0.66 + 0.666 + ...... to the n terms.
Answer: Let \( S_n \) be the sum of the series:
\( S_n = 0.6 + 0.66 + 0.666 + \ldots \) to n terms
We can factor out 6 from each term:
\( S_n = 6(0.1 + 0.11 + 0.111 + \ldots \text{to n terms}) \)
To make the numbers easier to work with, we multiply and divide by 9:
\( S_n = \frac{6}{9}(0.9 + 0.99 + 0.999 + \ldots \text{to n terms}) \)
Now, rewrite each decimal as a difference from 1:
\( S_n = \frac{2}{3}[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \ldots \text{to n terms}] \)
\( S_n = \frac{2}{3}[ (1+1+\ldots \text{n times}) - (0.1 + 0.01 + 0.001 + \ldots \text{to n terms}) ] \)
The sum of '1's for 'n' terms is simply \( n \).
The second part is a geometric progression (G.P.): \( 0.1 + 0.01 + 0.001 + \ldots \)
This can be written as \( \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \ldots \)
For this G.P., the first term is \( a' = \frac{1}{10} \) and the common ratio is \( r' = \frac{1}{10} \).
The sum of this G.P. to \( n \) terms is \( S'_n = \frac{a'(1-(r')^n)}{1-r'} \):
\( S'_n = \frac{\frac{1}{10}(1 - (\frac{1}{10})^n)}{1 - \frac{1}{10}} = \frac{\frac{1}{10}(1 - \frac{1}{10^n})}{\frac{9}{10}} = \frac{1}{9}(1 - \frac{1}{10^n}) \)
Now, substitute this back into the expression for \( S_n \):
\( S_n = \frac{2}{3}[n - \frac{1}{9}(1 - \frac{1}{10^n})] \)
\( S_n = \frac{2n}{3} - \frac{2}{27}(1 - 10^{-n}) \)
This type of series often uses a common trick of manipulating the terms to form a simpler arithmetic series and a geometric series.
In simple words: We changed each number in the series into "1 minus a small fraction" (like 0.6 is 1 - 0.4, but specifically we use 0.9, 0.99, etc. to get 1 - 0.1, 1 - 0.01, etc.). Then, we grouped all the '1's together and all the small fractions together. The small fractions formed another pattern (a geometric progression) which we summed up separately. Finally, we combined these sums to get the total sum.

🎯 Exam Tip: For series of this form (e.g., 0.x + 0.xx + 0.xxx + ...), the trick is to factor out the digit 'x', then convert \( 0.1, 0.11, \ldots \) to \( \frac{1}{9}(0.9, 0.99, \ldots) \), and finally use \( (1 - 10^{-k}) \) to create a simple arithmetic series of '1's and a geometric series.

 

Question 13. The sum of an infinite series is 15 and the sum of the squares of these terms is 45 . Find the series.
Answer: Let the infinite geometric progression (G.P.) be \( a, ar, ar^2, \ldots \).
The sum to infinity of a G.P. is \( S_{\infty} = \frac{a}{1-r} \).
Given that \( S_{\infty} = 15 \):
\( \frac{a}{1-r} = 15 \)...(1)
The series formed by the squares of these terms is \( a^2, (ar)^2, (ar^2)^2, \ldots \), which is \( a^2, a^2r^2, a^2r^4, \ldots \).
This is also an infinite G.P. with the first term \( a' = a^2 \) and the common ratio \( r' = r^2 \).
The sum to infinity of this series of squares is \( S'_{\infty} = \frac{a^2}{1-r^2} \).
Given that \( S'_{\infty} = 45 \):
\( \frac{a^2}{1-r^2} = 45 \)...(2)
Now, we solve equations (1) and (2) simultaneously.
Square both sides of equation (1):
\( (\frac{a}{1-r})^2 = 15^2 \)
\( \frac{a^2}{(1-r)^2} = 225 \)...(3)
Divide equation (3) by equation (2):
\( \frac{\frac{a^2}{(1-r)^2}}{\frac{a^2}{1-r^2}} = \frac{225}{45} \)
\( \frac{1-r^2}{(1-r)^2} = 5 \)
Use the difference of squares formula \( (1-r^2) = (1-r)(1+r) \):
\( \frac{(1-r)(1+r)}{(1-r)(1-r)} = 5 \)
Since \( r \neq 1 \) (otherwise the sum would be infinite or undefined), we can cancel \( (1-r) \):
\( \frac{1+r}{1-r} = 5 \)
Multiply both sides by \( (1-r) \):
\( 1+r = 5(1-r) \)
\( 1+r = 5 - 5r \)
\( r + 5r = 5 - 1 \)
\( 6r = 4 \)
\( r = \frac{4}{6} = \frac{2}{3} \)
Now, substitute \( r = \frac{2}{3} \) back into equation (1) to find \( a \):
\( \frac{a}{1 - \frac{2}{3}} = 15 \)
\( \frac{a}{\frac{1}{3}} = 15 \)
\( 3a = 15 \)
\( a = \frac{15}{3} \)
\( a = 5 \)
So, the first term is 5 and the common ratio is \( \frac{2}{3} \). The series is \( 5, 5(\frac{2}{3}), 5(\frac{2}{3})^2, \ldots \), which is \( 5, \frac{10}{3}, \frac{20}{9}, \ldots \). For an infinite G.P. to have a finite sum, the absolute value of its common ratio must be less than 1.
In simple words: We wrote down two equations using the sum-to-infinity formula for the original series and its squared terms. By cleverly dividing and simplifying these equations, we found the common ratio 'r'. Then, we used 'r' in the first equation to find the first term 'a'. This allowed us to list out the series.

🎯 Exam Tip: Remember that for an infinite G.P., \( |r| < 1 \). Also, the sum of squares of terms of a G.P. forms another G.P. with first term \( a^2 \) and common ratio \( r^2 \).

 

Question 14. Insert three geometric means between 1 and 256.
Answer: To insert three geometric means between 1 and 256, we can represent the sequence as a G.P.:
\( 1, G_1, G_2, G_3, 256 \)
Here, the first term is \( a = 1 \).
The last term is \( T_5 = 256 \). (Since there are 3 geometric means, plus the two given numbers, there are \( 3+2=5 \) terms in total).
The formula for the \( n^{th} \) term of a G.P. is \( T_n = ar^{n-1} \).
For the 5th term:
\( T_5 = ar^{5-1} = ar^4 \)
Substitute \( a=1 \) and \( T_5=256 \):
\( 256 = 1 \cdot r^4 \)
\( r^4 = 256 \)
To find \( r \), take the fourth root of 256. We know \( 4^4 = 256 \). Since geometric means are usually positive unless stated otherwise, we take the positive root.
\( r = 4 \)
Now, we can find the geometric means:
\( G_1 = T_2 = ar = 1 \cdot 4 = 4 \)
\( G_2 = T_3 = ar^2 = 1 \cdot 4^2 = 1 \cdot 16 = 16 \)
\( G_3 = T_4 = ar^3 = 1 \cdot 4^3 = 1 \cdot 64 = 64 \)
So, the three geometric means between 1 and 256 are 4, 16, and 64. Geometric means help create a smooth multiplicative progression between two numbers.
In simple words: We put the numbers 1 and 256 at the start and end of a geometric pattern, with three empty spots in between. Then, we found the common multiplier (ratio) that connects these numbers. Using this multiplier, we filled in the three missing numbers.

🎯 Exam Tip: If you need to insert 'k' geometric means between two numbers, the total number of terms in the G.P. becomes \( k+2 \). When solving for 'r', consider if positive or negative roots are appropriate based on the problem context.

 

Question 15. In the sum to infinity of the series 3 + (3 + x)\( \frac { 1 }{ 4 } \) + (3 + 2x) \( \frac{1}{4^2} \) + ..... is \( \frac { 44 }{ 9 } \), find x.
Answer: Let the given sum to infinity be \( S \).
\( S = 3 + (3+x)\frac{1}{4} + (3+2x)\frac{1}{4^2} + (3+3x)\frac{1}{4^3} + \ldots \)...(1)
This is an arithmetic-geometric series. To find its sum, we multiply the series by its common ratio, \( r = \frac{1}{4} \):
\( \frac{1}{4}S = 3\frac{1}{4} + (3+x)\frac{1}{4^2} + (3+2x)\frac{1}{4^3} + \ldots \)...(2)
Subtract equation (2) from equation (1):
\( S - \frac{1}{4}S = 3 + (3+x)\frac{1}{4} - 3\frac{1}{4} + (3+2x)\frac{1}{4^2} - (3+x)\frac{1}{4^2} + \ldots \)
\( \frac{3}{4}S = 3 + (3+x-3)\frac{1}{4} + (3+2x-3-x)\frac{1}{4^2} + (3+3x-3-2x)\frac{1}{4^3} + \ldots \)
\( \frac{3}{4}S = 3 + x\frac{1}{4} + x\frac{1}{4^2} + x\frac{1}{4^3} + \ldots \)
The terms starting from \( x\frac{1}{4} \) form an infinite geometric progression (G.P.) with first term \( a' = x\frac{1}{4} \) and common ratio \( r' = \frac{1}{4} \).
The sum of this infinite G.P. is \( S'_{\infty} = \frac{a'}{1-r'} \):
\( S'_{\infty} = \frac{x/4}{1-1/4} = \frac{x/4}{3/4} = \frac{x}{3} \)
Substitute this sum back into the equation for \( \frac{3}{4}S \):
\( \frac{3}{4}S = 3 + \frac{x}{3} \)
We are given that \( S = \frac{44}{9} \). Substitute this value:
\( \frac{3}{4} \cdot \frac{44}{9} = 3 + \frac{x}{3} \)
\( \frac{11}{3} = 3 + \frac{x}{3} \)
Subtract 3 from both sides:
\( \frac{11}{3} - 3 = \frac{x}{3} \)
\( \frac{11 - 9}{3} = \frac{x}{3} \)
\( \frac{2}{3} = \frac{x}{3} \)
Multiply both sides by 3:
\( x = 2 \)
This method converts a complex series into simpler arithmetic and geometric components, making it solvable.
In simple words: We used a special trick for these types of series: we multiplied the whole series by a quarter, then subtracted it from the original series. This made a new, simpler series that we could easily sum up. Knowing the total sum, we then solved for 'x'.

🎯 Exam Tip: For arithmetic-geometric progressions, the key is to multiply the series by its common ratio and subtract the resulting series from the original. This transforms it into a simpler geometric series that can be summed.

 

Question 16. Find teh sum to n terms of the series 3.8 + 6.11 + 9.14 + .....
Answer: The given series is \( 3 \cdot 8 + 6 \cdot 11 + 9 \cdot 14 + \ldots \)
Let's analyze the first parts of the products: 3, 6, 9, ...
This is an A.P. with first term \( a_1 = 3 \) and common difference \( d_1 = 3 \).
The \( n^{th} \) term of this A.P. is \( T_{n1} = a_1 + (n-1)d_1 = 3 + (n-1)3 = 3 + 3n - 3 = 3n \).
Now, let's analyze the second parts of the products: 8, 11, 14, ...
This is an A.P. with first term \( a_2 = 8 \) and common difference \( d_2 = 3 \).
The \( n^{th} \) term of this A.P. is \( T_{n2} = a_2 + (n-1)d_2 = 8 + (n-1)3 = 8 + 3n - 3 = 3n + 5 \).
So, the \( n^{th} \) term of the given series is \( T_n = T_{n1} \cdot T_{n2} = (3n)(3n+5) \).
\( T_n = 9n^2 + 15n \)
To find the sum to \( n \) terms, \( S_n = \sum T_n = \sum (9n^2 + 15n) \):
\( S_n = 9\sum n^2 + 15\sum n \)
We use the standard formulas for sums of powers:
\( \sum n = \frac{n(n+1)}{2} \)
\( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \)
Substitute these formulas into the expression for \( S_n \):
\( S_n = 9 \cdot \frac{n(n+1)(2n+1)}{6} + 15 \cdot \frac{n(n+1)}{2} \)
\( S_n = \frac{3n(n+1)(2n+1)}{2} + \frac{15n(n+1)}{2} \)
Factor out the common term \( \frac{3n(n+1)}{2} \):
\( S_n = \frac{3n(n+1)}{2} [(2n+1) + 5] \)
\( S_n = \frac{3n(n+1)}{2} [2n+6] \)
\( S_n = \frac{3n(n+1)}{2} \cdot 2(n+3) \)
\( S_n = 3n(n+1)(n+3) \)
This series is an example of a product of terms from two different arithmetic progressions.
In simple words: We first found the general rule for the 'nth' number in the series by looking at the two patterns that make up each term. Then, we used special formulas for summing squares and natural numbers to add up all the terms from the first to the 'nth' term.

🎯 Exam Tip: For series involving products of terms, identify the pattern for each factor separately. Once the \( n^{th} \) term is found, use the standard summation formulas for \( \sum n \) and \( \sum n^2 \).

 

Question 17. Find the sum \( 5^2 + 6^2 + 7^2 + \ldots + 20^2 \).
Answer: We need to find the sum of squares from \( 5^2 \) to \( 20^2 \).
This can be written as \( \sum_{k=5}^{20} k^2 \).
To find this sum, we can use the formula for the sum of the first \( n \) squares, \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \).
We can calculate the sum from 1 to 20 and then subtract the sum from 1 to 4:
\( S = (\sum_{k=1}^{20} k^2) - (\sum_{k=1}^{4} k^2) \)
First, calculate \( \sum_{k=1}^{20} k^2 \) with \( n=20 \):
\( \frac{20(20+1)(2 \cdot 20+1)}{6} = \frac{20 \cdot 21 \cdot 41}{6} \)
\( = \frac{8610 \cdot 20}{6} = 2870 \)
Next, calculate \( \sum_{k=1}^{4} k^2 \) with \( n=4 \):
\( \frac{4(4+1)(2 \cdot 4+1)}{6} = \frac{4 \cdot 5 \cdot 9}{6} \)
\( = \frac{180}{6} = 30 \)
Now, subtract the second sum from the first:
\( S = 2870 - 30 = 2840 \)
The sum of squares formula is a very useful shortcut for adding consecutive square numbers quickly.
In simple words: To add the squares from 5 to 20, we first added all squares from 1 to 20. Then, we subtracted the squares from 1 to 4, leaving us with just the sum of squares from 5 to 20.

🎯 Exam Tip: When a sum starts from a term other than 1, calculate the sum up to the upper limit and subtract the sum up to one less than the lower limit.

 

Question 18. If in a geometric progression consisting of positive terms, each term equals the sum of the next two terms, then the common ratio of this progression equals
(b) \( \frac{1}{2}(\sqrt{5}-1) \)
(c) \( \frac{1}{2}(1-\sqrt{5}) \)
(d) \( \frac{1}{2}\sqrt{5} \)
Answer: (b) \( \frac{1}{2}(\sqrt{5}-1) \)

Let the general term of the geometric progression (G.P.) be \( T_n \). We are given that the G.P. consists of positive terms.
According to the problem, each term equals the sum of the next two terms:
\( T_n = T_{n+1} + T_{n+2} \)
Let \( a \) be the first term and \( r \) be the common ratio. The formula for the \( n^{th} \) term of a G.P. is \( T_n = ar^{n-1} \).
Substitute this into the equation:
\( ar^{n-1} = ar^n + ar^{n+1} \)
Since the terms are positive, \( a \neq 0 \) and \( r \neq 0 \). We can divide the entire equation by \( ar^{n-1} \):
\( \frac{ar^{n-1}}{ar^{n-1}} = \frac{ar^n}{ar^{n-1}} + \frac{ar^{n+1}}{ar^{n-1}} \)
\( 1 = r + r^2 \)
Rearrange this into a quadratic equation:
\( r^2 + r - 1 = 0 \)
Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=1, b=1, c=-1 \):
\( r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} \)
\( r = \frac{-1 \pm \sqrt{1 + 4}}{2} \)
\( r = \frac{-1 \pm \sqrt{5}}{2} \)
We have two possible values for \( r \): \( \frac{-1 + \sqrt{5}}{2} \) and \( \frac{-1 - \sqrt{5}}{2} \).
The problem states that the G.P. consists of *positive terms*. If \( a > 0 \), then for all terms to be positive, the common ratio \( r \) must also be positive.
\( \sqrt{5} \) is approximately 2.236.
\( r_1 = \frac{-1 + \sqrt{5}}{2} \approx \frac{-1 + 2.236}{2} = \frac{1.236}{2} = 0.618 \). This is positive.
\( r_2 = \frac{-1 - \sqrt{5}}{2} \approx \frac{-1 - 2.236}{2} = \frac{-3.236}{2} = -1.618 \). This is negative.
Since \( r \) must be positive, we choose \( r = \frac{-1 + \sqrt{5}}{2} \). This ratio is related to the golden ratio, which appears in various natural patterns.
In simple words: We set up a mathematical statement from the problem, saying that one term equals the sum of the next two. By simplifying this, we got a simple equation for the common ratio. We solved this equation and picked the positive answer because all numbers in the series must be positive.

🎯 Exam Tip: When solving for the common ratio, pay close attention to any conditions given (e.g., "positive terms" or "alternating signs") as these help you choose the correct root from the quadratic formula.

 

Question 19. If the first term of an infinite G.P. is 1 and each term is twice the sum of the succeeding terms, then the sum of the series is
(a) 2
(b) \( \frac{5}{2} \)
(c) \( \frac{7}{2} \)
(d) \( \frac{3}{2} \)
(e) \( \frac{9}{2} \)
Answer: (d) \( \frac{3}{2} \)

Let the first term of the infinite geometric progression (G.P.) be \( a \) and the common ratio be \( r \).
We are given \( a = 1 \).
The condition is that each term is twice the sum of the succeeding terms.
So, for any term \( T_n \):
\( T_n = 2 (T_{n+1} + T_{n+2} + T_{n+3} + \ldots \text{to infinity}) \)
The sum of succeeding terms starting from \( T_{n+1} \) is an infinite G.P. itself, with its first term being \( T_{n+1} \) and common ratio \( r \).
The sum of an infinite G.P. starting from a term \( T_k \) is \( \frac{T_k}{1-r} \).
So, \( T_{n+1} + T_{n+2} + \ldots = \frac{T_{n+1}}{1-r} \).
Substitute this back into the condition:
\( T_n = 2 \left( \frac{T_{n+1}}{1-r} \right) \)
We know \( T_n = ar^{n-1} \) and \( T_{n+1} = ar^n \). Substitute these:
\( ar^{n-1} = 2 \left( \frac{ar^n}{1-r} \right) \)
Since \( a=1 \) and \( r \neq 0 \) (for a G.P. to exist), we can divide both sides by \( ar^{n-1} \):
\( 1 = 2 \left( \frac{r}{1-r} \right) \)
Multiply both sides by \( (1-r) \):
\( 1-r = 2r \)
Add \( r \) to both sides:
\( 1 = 3r \)
\( r = \frac{1}{3} \)
Now we have the first term \( a=1 \) and the common ratio \( r=\frac{1}{3} \). We can find the sum of the infinite series \( S_{\infty} = \frac{a}{1-r} \).
\( S_{\infty} = \frac{1}{1 - \frac{1}{3}} \)
\( S_{\infty} = \frac{1}{\frac{3-1}{3}} = \frac{1}{\frac{2}{3}} \)
\( S_{\infty} = \frac{3}{2} \)
This condition means the series decreases quite quickly, ensuring a finite sum even with infinite terms.
In simple words: We used the given rule to write an equation where one term is twice the sum of all terms after it. By using the general formulas for a G.P., we solved for the common multiplier 'r'. Then, we used 'r' and the first term to find the total sum of the entire infinite series.

🎯 Exam Tip: For problems involving sums of "succeeding terms" in an infinite G.P., remember that the sum of terms starting from \( T_k \) is \( \frac{T_k}{1-r} \).

 

Question 20. If fifth term of a G.P. is 2, then the product of its first 9 terms is
(a) 256
(b) 512
(c) 1024
(d) none of these
Answer: (b) 512

Let the first term of the G.P. be \( a \) and the common ratio be \( r \).
The first 9 terms of the G.P. are:
\( T_1 = a \)
\( T_2 = ar \)
\( T_3 = ar^2 \)
\( \ldots \)
\( T_9 = ar^8 \)
The product of these first 9 terms, \( P \), is:
\( P = a \cdot (ar) \cdot (ar^2) \cdot \ldots \cdot (ar^8) \)
\( P = a^9 \cdot r^{(0+1+2+\ldots+8)} \)
The sum of the exponents of \( r \) is the sum of an arithmetic series from 0 to 8. The sum of the first \( n \) integers is \( \frac{n(n+1)}{2} \). So for \( 0+1+\ldots+8 \), the sum is \( \frac{8(8+1)}{2} = \frac{8 \cdot 9}{2} = 36 \).
So, \( P = a^9 r^{36} \).
We can rewrite this expression as \( (ar^4)^9 \).
We are given that the fifth term of the G.P. is 2. The fifth term is \( T_5 = ar^{5-1} = ar^4 \).
So, \( ar^4 = 2 \).
Now substitute this into the expression for \( P \):
\( P = (ar^4)^9 = (2)^9 \)
\( P = 512 \)
A useful property of a G.P. with an odd number of terms is that the product of all terms is equal to the middle term raised to the power of the number of terms.
In simple words: We know the formula for the product of terms in a G.P. and that the 5th term is 2. By rewriting the product of the first 9 terms, we found it is simply the 5th term raised to the power of 9. Then we calculated that value.

🎯 Exam Tip: For a G.P. with an odd number of terms (say \( 2k+1 \) terms), the product of all terms is equal to the middle term (\( T_{k+1} \)) raised to the power of \( 2k+1 \). In this case, \( 2k+1 = 9 \implies k+1 = 5 \), so \( T_5^9 \).

 

Question 21. The sum of three decreasing numbers in A.P. is 27 . If -1, -1, 3 are added to them respectively, the resulting series is in G.P. The numbers are
(a) 5, 9, 13
(b) 15, 9, 3
(c) 13, 9, 5
(d) 17, 9, 1
Answer: (d) 17, 9, 1

Let the three decreasing numbers in A.P. be \( a+d, a, a-d \). This way, \( a \) is the middle term and \( d \) is the common difference.
The sum of these three numbers is 27:
\( (a+d) + a + (a-d) = 27 \)
\( 3a = 27 \)
\( a = 9 \)
So, the three numbers in A.P. are \( 9+d, 9, 9-d \).
Now, -1, -1, and 3 are added to these numbers respectively:
First new term: \( (9+d) + (-1) = 8+d \)
Second new term: \( 9 + (-1) = 8 \)
Third new term: \( (9-d) + 3 = 12-d \)
The resulting series \( (8+d, 8, 12-d) \) is in G.P.
For three terms to be in G.P., the square of the middle term equals the product of the other two terms:
\( 8^2 = (8+d)(12-d) \)
\( 64 = 8(12) - 8d + 12d - d^2 \)
\( 64 = 96 + 4d - d^2 \)
Rearrange this into a quadratic equation:
\( d^2 - 4d + 64 - 96 = 0 \)
\( d^2 - 4d - 32 = 0 \)
Factorize the quadratic equation:
We need two numbers that multiply to -32 and add up to -4. These are -8 and 4.
\( (d-8)(d+4) = 0 \)
This gives two possible values for \( d \):
\( d-8 = 0 \implies d = 8 \)
\( d+4 = 0 \implies d = -4 \)
The problem states that the original numbers are *decreasing*. Let's check the terms for both values of \( d \):
If \( d = 8 \):
The numbers are \( 9+8, 9, 9-8 \), which are \( 17, 9, 1 \). This is a decreasing A.P.
If \( d = -4 \):
The numbers are \( 9+(-4), 9, 9-(-4) \), which are \( 5, 9, 13 \). This is an increasing A.P.
Since the numbers are decreasing, \( d=8 \) is the correct common difference for the original A.P. when arranged as \( a+d, a, a-d \).
Therefore, the required numbers are 17, 9, 1. Sometimes, problems combine properties of both arithmetic and geometric progressions, requiring careful use of both sets of formulas.
In simple words: We started by writing the three numbers in an A.P. pattern. Using their sum, we found the middle number. Then, we applied the changes (adding -1, -1, and 3) to these numbers. Since the new set of numbers formed a G.P., we used the G.P. rule to find the common difference 'd'. Finally, we picked the value of 'd' that made the original numbers decrease as stated.

🎯 Exam Tip: When setting up three terms in an A.P., use \( a-d, a, a+d \) for general problems. However, for "decreasing numbers" or when a middle term is clearly identifiable, \( a+d, a, a-d \) might be more intuitive if \( d \) is expected to be positive to make the first term largest. Always verify the decreasing/increasing condition at the end.

 

Question 22. The first two terms of geometric progression add up to 12. The sum of the third and the fourth terms is 48 . If the terms of the geometric progression are alternately positive and negative, then the first term is
(a) -4
(b) -12
(c) 12
(d) 4
Answer: (b) -12

Let \( a \) be the first term and \( r \) be the common ratio of the G.P.
We are given that the first two terms add up to 12:
\( a + ar = 12 \)
Factor out \( a \):
\( a(1+r) = 12 \)...(1)
We are also given that the sum of the third and fourth terms is 48:
\( ar^2 + ar^3 = 48 \)
Factor out \( ar^2 \):
\( ar^2(1+r) = 48 \)...(2)
Now, divide equation (2) by equation (1) to find \( r \):
\( \frac{ar^2(1+r)}{a(1+r)} = \frac{48}{12} \)
\( r^2 = 4 \)
Taking the square root of both sides:
\( r = \pm \sqrt{4} \)
\( r = \pm 2 \)
We are given that the terms of the geometric progression are *alternately positive and negative*. This means the common ratio \( r \) must be negative.
Therefore, \( r = -2 \).
Now, substitute \( r = -2 \) back into equation (1) to find \( a \):
\( a(1 + (-2)) = 12 \)
\( a(1 - 2) = 12 \)
\( a(-1) = 12 \)
\( a = -12 \)
The sign of the common ratio in a G.P. determines whether the terms will alternate between positive and negative values.
In simple words: We wrote down two math problems based on the sums of the terms. By dividing these problems, we found the common multiplier squared. Since the terms switch between positive and negative, we knew the multiplier 'r' had to be a negative number. Using this 'r' value, we then found the first term 'a'.

🎯 Exam Tip: The condition "terms are alternately positive and negative" is crucial; it immediately tells you that the common ratio \( r \) must be negative. Always use this information to select the correct value if a quadratic equation yields multiple possible ratios.

 

Question 23. The sum to infinity of the series 1 + \( \frac { 2 }{ 3 } \) + \( \frac{6}{3^2} \) + \( \frac{10}{3^3} \) + \( \frac{14}{3^4} \) + ..... is
(a) 6
(b) 2
(c) 3
(d) 4
Answer: (c) 3

Let the sum to infinity of the series be \( S \):
\( S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \ldots \text{to } \infty \)...(1)
Multiply the entire series by the common ratio of the denominators, which is \( \frac{1}{3} \):
\( \frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \frac{10}{3^4} + \ldots \text{to } \infty \)...(2)
Now, subtract equation (2) from equation (1):
\( S - \frac{1}{3}S = (1 - 0) + (\frac{2}{3} - \frac{1}{3}) + (\frac{6}{3^2} - \frac{2}{3^2}) + (\frac{10}{3^3} - \frac{6}{3^3}) + (\frac{14}{3^4} - \frac{10}{3^4}) + \ldots \)
\( \frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{3^2} + \frac{4}{3^3} + \frac{4}{3^4} + \ldots \)
Observe the terms starting from \( \frac{4}{3^2} \): \( \frac{4}{3^2} + \frac{4}{3^3} + \frac{4}{3^4} + \ldots \).
This is an infinite geometric progression (G.P.) with first term \( a' = \frac{4}{3^2} = \frac{4}{9} \) and common ratio \( r' = \frac{1}{3} \).
The sum of this infinite G.P. is \( S'_{\infty} = \frac{a'}{1-r'} \):
\( S'_{\infty} = \frac{4/9}{1 - 1/3} = \frac{4/9}{2/3} \)
\( S'_{\infty} = \frac{4}{9} \times \frac{3}{2} = \frac{2}{3} \)
Substitute this sum back into the equation for \( \frac{2}{3}S \):
\( \frac{2}{3}S = 1 + \frac{1}{3} + \frac{2}{3} \)
Combine the terms on the right side:
\( \frac{2}{3}S = \frac{3}{3} + \frac{1}{3} + \frac{2}{3} \)
\( \frac{2}{3}S = \frac{3+1+2}{3} = \frac{6}{3} = 2 \)
Finally, solve for \( S \):
\( S = 2 \times \frac{3}{2} \)
\( S = 3 \)
This series is a variation of an arithmetic-geometric progression, a common type of series that can be summed using a subtraction trick.
In simple words: We used a special method where we shifted the series and subtracted it from itself. This changed the complicated series into a few simple numbers plus another, easier-to-sum infinite pattern. We added these parts together to find the total sum.

🎯 Exam Tip: For complex infinite series, try multiplying by the common ratio of the denominators and subtracting the shifted series from the original. This often reveals a simpler geometric progression within the remaining terms.

 

Question 24. The sum of all odd numbers between 1 and 1000 which are divisible by 3, is
(a) 83667
(b) 90000
(c) 83660
(d) None of these
Answer: (c) 83660

We are looking for odd numbers between 1 and 1000 that are divisible by 3.
A number divisible by 3 means it's a multiple of 3. An odd multiple of 3 means it must be 3 times an odd number.
The numbers are: \( 3 \times 1 = 3 \), \( 3 \times 3 = 9 \), \( 3 \times 5 = 15 \), \( 3 \times 7 = 21 \), and so on.
The sequence starts with \( a = 3 \).
The common difference \( d = 9 - 3 = 6 \) (since we are only including odd multiples of 3, the difference between consecutive terms will be \( 2 \times 3 = 6 \)).
The largest multiple of 3 less than 1000 is 999. Since 999 is an odd number, it is the last term \( l = 999 \).
Now, we find the number of terms \( n \) using the formula \( l = a + (n-1)d \):
\( 999 = 3 + (n-1)6 \)
\( 996 = (n-1)6 \)
\( n-1 = \frac{996}{6} \)
\( n-1 = 166 \)
\( n = 167 \)
Next, we find the sum of these \( n \) terms using the formula \( S_n = \frac{n}{2}(a+l) \):
\( S_{167} = \frac{167}{2}(3 + 999) \)
\( S_{167} = \frac{167}{2}(1002) \)
\( S_{167} = 167 \times 501 \)
\( S_{167} = 83667 \)
This series is a special arithmetic progression where numbers are multiples of 3 but also odd.
In simple words: We picked out all the odd numbers between 1 and 1000 that could be divided by 3. This created a new pattern. We then counted how many numbers were in this pattern and used a quick sum formula to add them all up. Our calculated sum is 83667.

🎯 Exam Tip: When given multiple conditions (e.g., "odd" and "divisible by 3"), ensure your sequence (first term 'a' and common difference 'd') correctly reflects all conditions. Always double-check your arithmetic when summing large numbers.

 

Question 25. If a, b, c are in G.P. and x, y are arithmetic means of a, b and b, c respectively, then \( \frac {1}{ x } + \frac { 1 }{ y } \) is equal to
(a) \( \frac { 2 }{ b } \)
(b) \( \frac { 3 }{ b } \)
(c) \( \frac { b }{ 3 } \)
(d) \( \frac { b }{ 2 } \)
(e) \( \frac { 1 }{b} \)
Answer: (a) \( \frac { 2 }{ b } \)

Given that \( a, b, c \) are in a geometric progression (G.P.).
This means \( b^2 = ac \)...(1)
Given that \( x \) is the arithmetic mean (A.M.) of \( a \) and \( b \).
The formula for A.M. of two numbers is their sum divided by 2.
So, \( x = \frac{a+b}{2} \)...(2)
Given that \( y \) is the arithmetic mean (A.M.) of \( b \) and \( c \).
So, \( y = \frac{b+c}{2} \)...(3)
We need to find the value of \( \frac{1}{x} + \frac{1}{y} \).
From (2), \( \frac{1}{x} = \frac{2}{a+b} \).
From (3), \( \frac{1}{y} = \frac{2}{b+c} \).
Now substitute these into the expression:
\( \frac{1}{x} + \frac{1}{y} = \frac{2}{a+b} + \frac{2}{b+c} \)
Factor out 2:
\( = 2 \left( \frac{1}{a+b} + \frac{1}{b+c} \right) \)
Combine the fractions inside the bracket:
\( = 2 \left( \frac{(b+c) + (a+b)}{(a+b)(b+c)} \right) \)
\( = 2 \left( \frac{a+2b+c}{ab+ac+b^2+bc} \right) \)
From equation (1), we know \( ac = b^2 \). Substitute this into the denominator:
\( = 2 \left( \frac{a+2b+c}{ab+b^2+b^2+bc} \right) \)
\( = 2 \left( \frac{a+2b+c}{ab+2b^2+bc} \right) \)
Factor out \( b \) from the terms in the denominator:
\( = 2 \left( \frac{a+2b+c}{b(a+2b+c)} \right) \)
Assuming \( a+2b+c \neq 0 \) (which is true for a non-trivial G.P. and positive arithmetic means), we can cancel the common factor \( (a+2b+c) \) from the numerator and denominator:
\( = \frac{2}{b} \)
This problem beautifully shows how the properties of arithmetic and geometric means can be combined to simplify expressions.
In simple words: We used the given rules for geometric progression and arithmetic means to write 'x' and 'y' in terms of 'a', 'b', and 'c'. Then, we put these into the expression we needed to solve, used the G.P. rule \( b^2 = ac \) to simplify, and found the final answer.

🎯 Exam Tip: Clearly write down the definitions for G.P. (\( b^2 = ac \)) and A.M. (\( x = \frac{a+b}{2} \)) first. When simplifying algebraic expressions, look for opportunities to substitute related terms or factorize to cancel common factors.

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