Samacheer Kalvi Class 10 Maths Solutions Chapter 2 Numbers and Sequences Exercise 2.8

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers And Sequences Ex 2.8

Question 1. Find the sum of first n terms of the G.P.
(i) 5, -3, \( \frac { 9 }{ 5 } \), \( -\frac { 27 }{ 25 } \),...
(ii) 256,64,16,.......
Answer:
(i) Given G.P. is 5, -3, \( \frac { 9 }{ 5 } \), \( -\frac { 27 }{ 25 } \),... up to n terms.
Here, the first term \( a = 5 \).
The common ratio \( r = \frac { -3 }{ 5 } \).
Since \( |r| = |-\frac { 3 }{ 5 }| = \frac { 3 }{ 5 } < 1 \), we use the sum to n terms formula:
\( S_n = \frac { a(1-r^n) }{ 1-r } \)
Substitute the values:
\( S_n = \frac { 5(1-(-\frac { 3 }{ 5 })^n) }{ 1-(-\frac { 3 }{ 5 }) } \)
\( \implies S_n = \frac { 5(1-(-\frac { 3 }{ 5 })^n) }{ 1+\frac { 3 }{ 5 } } \)
\( \implies S_n = \frac { 5(1-(-\frac { 3 }{ 5 })^n) }{ \frac { 5+3 }{ 5 } } \)
\( \implies S_n = \frac { 5(1-(-\frac { 3 }{ 5 })^n) }{ \frac { 8 }{ 5 } } \)
\( \implies S_n = 5 \times \frac { 5 }{ 8 } (1-(-\frac { 3 }{ 5 })^n) \)
\( \implies S_n = \frac { 25 }{ 8 } (1-(-\frac { 3 }{ 5 })^n) \)

(ii) Given G.P. is 256, 64, 16,.......
Here, the first term \( a = 256 \).
The common ratio \( r = \frac { 64 }{ 256 } = \frac { 1 }{ 4 } \).
Since \( |r| = |\frac { 1 }{ 4 }| = \frac { 1 }{ 4 } < 1 \), we use the sum to n terms formula:
\( S_n = \frac { a(1-r^n) }{ 1-r } \)
Substitute the values:
\( S_n = \frac { 256(1-(\frac { 1 }{ 4 })^n) }{ 1-\frac { 1 }{ 4 } } \)
\( \implies S_n = \frac { 256(1-(\frac { 1 }{ 4 })^n) }{ \frac { 4-1 }{ 4 } } \)
\( \implies S_n = \frac { 256(1-(\frac { 1 }{ 4 })^n) }{ \frac { 3 }{ 4 } } \)
\( \implies S_n = 256 \times \frac { 4 }{ 3 } (1-(\frac { 1 }{ 4 })^n) \)
\( \implies S_n = \frac { 1024 }{ 3 } (1-(\frac { 1 }{ 4 })^n) \)
In simple words: For each geometric progression, we find the first term 'a' and the common ratio 'r'. Then we use the correct formula for the sum of 'n' terms, depending on if 'r' is greater or smaller than 1. Make sure to carefully calculate the fractions.

🎯 Exam Tip: Always check the value of the common ratio 'r' to determine which formula for the sum of 'n' terms (\( S_n \)) to use: \( \frac { a(r^n-1) }{ r-1 } \) if \( r > 1 \) or \( \frac { a(1-r^n) }{ 1-r } \) if \( r < 1 \).

Question 2. Find the sum of first six terms of the G.P. 5,15,45,...
Answer:
Given G.P. is 5, 15, 45,...
Here, the first term \( a = 5 \).
The common ratio \( r = \frac { 15 }{ 5 } = 3 \).
The number of terms \( n = 6 \).
Since \( r = 3 > 1 \), we use the sum to n terms formula:
\( S_n = \frac { a(r^n-1) }{ r-1 } \)
Substitute the values:
\( S_6 = \frac { 5(3^6-1) }{ 3-1 } \)
\( \implies S_6 = \frac { 5(729-1) }{ 2 } \)
\( \implies S_6 = \frac { 5 \times 728 }{ 2 } \)
\( \implies S_6 = 5 \times 364 \)
\( \implies S_6 = 1820 \)
The sum of the first 6 terms is 1820.
In simple words: We found the first term and the common ratio. Since the common ratio was bigger than 1, we used a specific formula to add up the first six numbers in the series, which gave us 1820.

🎯 Exam Tip: When calculating sums, compute \( r^n \) accurately first, especially for larger exponents, to avoid arithmetic errors.

Question 3. Find the first term of the G.P. whose common ratio 5 and whose sum to first 6 terms is 46872.
Answer:
Given common ratio \( r = 5 \).
Given sum to first 6 terms \( S_6 = 46872 \).
The number of terms \( n = 6 \).
Since \( r = 5 > 1 \), we use the sum to n terms formula:
\( S_n = \frac { a(r^n-1) }{ r-1 } \)
Substitute the given values into the formula:
\( 46872 = \frac { a(5^6-1) }{ 5-1 } \)
\( \implies 46872 = \frac { a(15625-1) }{ 4 } \)
\( \implies 46872 = \frac { a(15624) }{ 4 } \)
Now, we solve for \( a \):
\( a = \frac { 46872 \times 4 }{ 15624 } \)
\( \implies a = \frac { 187488 }{ 15624 } \)
\( \implies a = 12 \)
The first term of the G.P. is 12.
In simple words: We knew the common ratio, the number of terms, and the total sum. We used the sum formula to work backward and find the first number in the pattern, which turned out to be 12.

🎯 Exam Tip: When solving for an unknown term like 'a', carefully isolate it by performing inverse operations. Double-check calculations for multiplication and division.

Question 4. Find the sum to infinity of (i) 9 + 3 + 1 + ....(ii) 21 + 14 + \( \frac { 28 }{ 3 } \) ......
Answer:
(i) Given G.P. is 9 + 3 + 1 + ....
Here, the first term \( a = 9 \).
The common ratio \( r = \frac { 3 }{ 9 } = \frac { 1 }{ 3 } \).
Since \( |r| = \frac { 1 }{ 3 } < 1 \), the sum to infinity \( S_\infty \) exists and is given by:
\( S_\infty = \frac { a }{ 1-r } \)
Substitute the values:
\( S_\infty = \frac { 9 }{ 1-\frac { 1 }{ 3 } } \)
\( \implies S_\infty = \frac { 9 }{ \frac { 3-1 }{ 3 } } \)
\( \implies S_\infty = \frac { 9 }{ \frac { 2 }{ 3 } } \)
\( \implies S_\infty = 9 \times \frac { 3 }{ 2 } \)
\( \implies S_\infty = \frac { 27 }{ 2 } \)

(ii) Given G.P. is 21 + 14 + \( \frac { 28 }{ 3 } \) ......
Here, the first term \( a = 21 \).
The common ratio \( r = \frac { 14 }{ 21 } = \frac { 2 }{ 3 } \).
Since \( |r| = \frac { 2 }{ 3 } < 1 \), the sum to infinity \( S_\infty \) exists and is given by:
\( S_\infty = \frac { a }{ 1-r } \)
Substitute the values:
\( S_\infty = \frac { 21 }{ 1-\frac { 2 }{ 3 } } \)
\( \implies S_\infty = \frac { 21 }{ \frac { 3-2 }{ 3 } } \)
\( \implies S_\infty = \frac { 21 }{ \frac { 1 }{ 3 } } \)
\( \implies S_\infty = 21 \times 3 \)
\( \implies S_\infty = 63 \)
In simple words: When numbers in a pattern get smaller and smaller by multiplying with a fraction, we can add them all up, even if they go on forever. This sum is called the sum to infinity. We use a special formula with the first term and the common ratio to find it.

🎯 Exam Tip: Remember that the sum to infinity of a G.P. exists only if the absolute value of the common ratio \( |r| \) is less than 1. Always check this condition first.

Question 5. If the first term of an infinite G.P. is 8 and its sum to infinity is \( \frac { 32 }{ 3 } \) then find the common ratio.
Answer:
Given first term \( a = 8 \).
Given sum to infinity \( S_\infty = \frac { 32 }{ 3 } \).
The formula for the sum to infinity of a G.P. is:
\( S_\infty = \frac { a }{ 1-r } \)
Substitute the given values:
\( \frac { 32 }{ 3 } = \frac { 8 }{ 1-r } \)
Now, we solve for \( r \):
\( 32(1-r) = 8 \times 3 \)
\( \implies 32 - 32r = 24 \)
\( \implies -32r = 24 - 32 \)
\( \implies -32r = -8 \)
\( \implies r = \frac { -8 }{ -32 } \)
\( \implies r = \frac { 1 }{ 4 } \)
The common ratio is \( \frac { 1 }{ 4 } \). This value is less than 1, so an infinite sum is possible.
In simple words: We used the formula for adding up an infinite series. We knew the first number and the total sum, so we worked backward to find the common fraction that each number was multiplied by to get the next.

🎯 Exam Tip: When solving for 'r', be careful with algebraic manipulations, especially when dealing with negative signs. Ensure your final 'r' value is consistent with the condition \( |r| < 1 \) for an infinite sum to exist.

Question 6. Find the sum to n terms of the series (i) 0.4 + 0.44 + 0.444 + ...... to n terms (ii) 3 + 33 + 333 + ...... to n terms
Answer:
(i) Given series is \( S_n = 0.4 + 0.44 + 0.444 + ...... \) to n terms.
We can rewrite the series:
\( S_n = 4(0.1 + 0.11 + 0.111 + ...... \text{ to n terms}) \)
Multiply and divide by 9:
\( S_n = \frac { 4 }{ 9 } (0.9 + 0.99 + 0.999 + ...... \text{ to n terms}) \)
Rewrite each term using 1 - (fraction):
\( S_n = \frac { 4 }{ 9 } [(1-\frac { 1 }{ 10 }) + (1-\frac { 1 }{ 100 }) + (1-\frac { 1 }{ 1000 }) + ...... \text{ to n terms}] \)
Separate the 1s and the fractions:
\( S_n = \frac { 4 }{ 9 } [(1+1+1+......\text{ (n times)}) - (\frac { 1 }{ 10 } + \frac { 1 }{ 100 } + \frac { 1 }{ 1000 } + ...... \text{ to n terms})] \)
The first part is \( n \). The second part is a G.P. with \( a = \frac { 1 }{ 10 } \) and \( r = \frac { 1 }{ 10 } \).
The sum of this G.P. is \( \frac { a(1-r^n) }{ 1-r } \):
\( S_n \text{ (G.P. part)} = \frac { \frac { 1 }{ 10 }(1-(\frac { 1 }{ 10 })^n) }{ 1-\frac { 1 }{ 10 } } \)
\( \implies S_n \text{ (G.P. part)} = \frac { \frac { 1 }{ 10 }(1-\frac { 1 }{ 10^n }) }{ \frac { 9 }{ 10 } } \)
\( \implies S_n \text{ (G.P. part)} = \frac { 1 }{ 10 } \times \frac { 10 }{ 9 } (1-\frac { 1 }{ 10^n }) \)
\( \implies S_n \text{ (G.P. part)} = \frac { 1 }{ 9 } (1-\frac { 1 }{ 10^n }) \)
Now substitute this back into the original \( S_n \) expression:
\( S_n = \frac { 4 }{ 9 } [n - \frac { 1 }{ 9 } (1-\frac { 1 }{ 10^n })] \)
\( \implies S_n = \frac { 4n }{ 9 } - \frac { 4 }{ 81 } (1-\frac { 1 }{ 10^n }) \)

(ii) Given series is \( S_n = 3 + 33 + 333 + ...... \) to n terms.
We can rewrite the series:
\( S_n = 3(1 + 11 + 111 + ...... \text{ to n terms}) \)
Multiply and divide by 9:
\( S_n = \frac { 3 }{ 9 } (9 + 99 + 999 + ...... \text{ to n terms}) \)
\( \implies S_n = \frac { 1 }{ 3 } [(10-1) + (100-1) + (1000-1) + ...... \text{ to n terms}] \)
Separate the powers of 10 and the -1s:
\( S_n = \frac { 1 }{ 3 } [(10+100+1000+...... \text{ to n terms}) - (1+1+1+...... \text{ (n times)})] \)
The first part is a G.P. with \( a = 10 \) and \( r = 10 \). The sum is \( \frac { a(r^n-1) }{ r-1 } \).
The second part is \( n \).
\( S_n = \frac { 1 }{ 3 } [\frac { 10(10^n-1) }{ 10-1 } - n] \)
\( \implies S_n = \frac { 1 }{ 3 } [\frac { 10(10^n-1) }{ 9 } - n] \)
\( \implies S_n = \frac { 10 }{ 27 } (10^n-1) - \frac { n }{ 3 } \)
In simple words: For these types of series, we first rewrite each term using fractions or powers of 10 to make them look like a geometric progression. Then, we use the formula for the sum of 'n' terms of a G.P. and simplify the expression to get the final answer.

🎯 Exam Tip: When dealing with repeating digit series like 0.444... or 333..., the key is to factor out the repeating digit and then convert the series into the form \( \frac { 1 }{ 9 } (10^n-1) \text{ or } (1 - \frac { 1 }{ 10^n }) \) to simplify the summation.

Question 7. Find the sum of the Geometric series 3 + 6 + 12 + ...... + 1536
Answer:
Given Geometric series is 3 + 6 + 12 + ...... + 1536.
Here, the first term \( a = 3 \).
The common ratio \( r = \frac { 6 }{ 3 } = 2 \).
The last term \( t_n = 1536 \).
First, we need to find the number of terms \( n \). The formula for the nth term of a G.P. is:
\( t_n = ar^{n-1} \)
Substitute the values:
\( 1536 = 3 \times 2^{n-1} \)
Divide both sides by 3:
\( \frac { 1536 }{ 3 } = 2^{n-1} \)
\( \implies 512 = 2^{n-1} \)
We know that \( 512 = 2^9 \). So,
\( 2^9 = 2^{n-1} \)
Equating the exponents:
\( 9 = n-1 \)
\( \implies n = 9+1 \)
\( \implies n = 10 \)
Now that we have the number of terms \( n = 10 \), we can find the sum of the series. Since \( r = 2 > 1 \), we use the formula:
\( S_n = \frac { a(r^n-1) }{ r-1 } \)
Substitute the values:
\( S_{10} = \frac { 3(2^{10}-1) }{ 2-1 } \)
\( \implies S_{10} = \frac { 3(1024-1) }{ 1 } \)
\( \implies S_{10} = 3 \times 1023 \)
\( \implies S_{10} = 3069 \)
The sum of the series is 3069. It is important to find the number of terms first when the last term is given.
In simple words: We first figured out how many numbers were in the pattern by using the last number given. Once we knew the count, we used the sum formula to add all the numbers together, which came out to 3069.

🎯 Exam Tip: When the last term of a G.P. is given instead of 'n', always start by calculating 'n' using the \( t_n = ar^{n-1} \) formula before finding the sum. This prevents calculation errors.

Question 8. Kumar writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs Rs. 2 to mail one letter, find the amount spent on postage when 8th set of letters is mailed.
Answer:
Kumar starts by mailing 4 letters. Each of those friends mails 4 letters, so 4x4 = 16 letters are sent next. This pattern forms a Geometric Progression (G.P.).
The G.P. is 4, 16, 64,.........
Here, the first term \( a = 4 \).
The common ratio \( r = \frac { 16 }{ 4 } = 4 \).
We need to find the total number of letters mailed up to the 8th set, which means finding the sum of the first 8 terms (\( S_8 \)).
Since \( r = 4 > 1 \), we use the sum to n terms formula:
\( S_n = \frac { a(r^n-1) }{ r-1 } \)
Substitute \( n = 8 \):
\( S_8 = \frac { 4(4^8-1) }{ 4-1 } \)
\( \implies S_8 = \frac { 4(65536-1) }{ 3 } \)
\( \implies S_8 = \frac { 4 \times 65535 }{ 3 } \)
\( \implies S_8 = \frac { 262140 }{ 3 } \)
\( \implies S_8 = 87380 \)
So, a total of 87380 letters are mailed by the 8th set. To find the total cost, we multiply by the cost per letter. This kind of chain letter shows how quickly numbers can grow in a geometric sequence.
Cost to mail one letter = Rs. 2.
Total amount spent on postage = Total letters mailed \( \times \) Cost per letter
Total amount spent = \( 87380 \times 2 \)
Total amount spent = Rs. 174760.
In simple words: Kumar's letter plan grows like a chain reaction. We found the total number of letters sent by the 8th round. Then we multiplied that number by the cost of sending one letter to get the total money spent.

🎯 Exam Tip: Recognize that this scenario describes a geometric progression. Carefully identify 'a', 'r', and 'n', then apply the correct sum formula and finally multiply by the cost per item.

Question 9. Find the rational form of the number 0.123 .
Answer:
Let \( x = 0.\overline{123} \). This means the digits 123 repeat infinitely.
So, \( x = 0.123123123.... \) --- (1)
Since there are 3 repeating digits, multiply both sides by \( 10^3 = 1000 \):
\( 1000x = 123.123123123.... \) --- (2)
Subtract equation (1) from equation (2):
\( 1000x - x = 123.123123... - 0.123123... \)
\( \implies 999x = 123 \)
\( \implies x = \frac { 123 }{ 999 } \)
This fraction can be simplified by dividing both the numerator and denominator by their greatest common divisor, which is 3.
\( x = \frac { 123 \div 3 }{ 999 \div 3 } \)
\( \implies x = \frac { 41 }{ 333 } \)
The rational form of \( 0.\overline{123} \) is \( \frac { 41 }{ 333 } \). Every repeating decimal can be written as a simple fraction.
In simple words: To change a repeating decimal into a fraction, we set the decimal equal to 'x', then multiply it by a power of 10 to shift the repeating part. Subtracting the original number from this new one helps us get rid of the repeating digits, leaving a simple equation to solve for 'x'.

🎯 Exam Tip: When converting repeating decimals to fractions, count the number of repeating digits to determine the correct power of 10 to multiply by. Always simplify the final fraction to its lowest terms.

Question 10. If \( S_n = (x + y) + (x^2 + xy + y^2) + (x^3 + x^2y + xy^2 + y^3) + ....... \) n terms then prove that \( (x-y)S_n = \frac { x^2(x^n-1) }{ x-1 } - \frac { y^2(y^n-1) }{ y-1 } \)
Answer:
Given the series \( S_n = (x + y) + (x^2 + xy + y^2) + (x^3 + x^2y + xy^2 + y^3) + ....... \) to n terms.
First, multiply \( S_n \) by \( (x-y) \). This makes each term simpler.
\( (x-y)S_n = (x-y)[(x+y) + (x^2+xy+y^2) + (x^3+x^2y+xy^2+y^3) + ....... \text{ to n terms}] \)
Recall the algebraic identities:
\( (x-y)(x+y) = x^2-y^2 \)
\( (x-y)(x^2+xy+y^2) = x^3-y^3 \)
\( (x-y)(x^3+x^2y+xy^2+y^3) = x^4-y^4 \)
Applying these identities to each term in the series:
\( (x-y)S_n = (x^2-y^2) + (x^3-y^3) + (x^4-y^4) + ....... \) to n terms.
Now, group the terms with \( x \) and \( y \) separately:
\( (x-y)S_n = (x^2+x^3+x^4+......\text{ to n terms}) - (y^2+y^3+y^4+......\text{ to n terms}) \)
Each of these grouped series is a Geometric Progression.
For the first G.P. (terms in x):
First term \( a_x = x^2 \).
Common ratio \( r_x = x \).
Number of terms is n.
Sum of this G.P. = \( \frac { a_x(r_x^n-1) }{ r_x-1 } = \frac { x^2(x^n-1) }{ x-1 } \)
For the second G.P. (terms in y):
First term \( a_y = y^2 \).
Common ratio \( r_y = y \).
Number of terms is n.
Sum of this G.P. = \( \frac { a_y(r_y^n-1) }{ r_y-1 } = \frac { y^2(y^n-1) }{ y-1 } \)
Substitute these sums back into the expression for \( (x-y)S_n \):
\( (x-y)S_n = \frac { x^2(x^n-1) }{ x-1 } - \frac { y^2(y^n-1) }{ y-1 } \)
Thus, the given statement is proved. This method is often used to simplify complex series into more manageable geometric series.
In simple words: We took the given series and multiplied it by \( (x-y) \). This made each part of the series turn into a simpler difference of powers. Then, we saw that these new series were actually two separate geometric progressions, and we used their sum formulas to prove the given equation.

🎯 Exam Tip: Recognize common algebraic factorizations like \( x^k - y^k = (x-y)(x^{k-1} + x^{k-2}y + ... + y^{k-1}) \). This identity is crucial for simplifying each term in the series before summing them up.