Solution: Given that, The sum of G.P of 3 terms is 125 Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ $125=\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)...
The common ratio of a G.P. is 3, and the last term is 486. If the sum of these terms be 728, find the first term.
Solution: Given that, The sum of GP $= 728$ Where, $r = 3, a = ?$ Firstly, $T_{n}=a r^{n-1}$ $486=a 3^{n-1}$ $486=a 3^{n} / 3$ $486(3)=a 3^{n}$ $1458=a 3^{n} \ldots .$ Eq. (i) $486=a 3^{n} / 3$...
The sum of n terms of the G.P. 3, 6, 12, … is 381. Find the value of n.
Solution: Given that, The sum of GP $=381$ Where, $a=3, r=6 / 3=2, n=?$ Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ $\begin{array}{l} 381=3\left(2^{n}-1\right)...
How many terms of the sequence √3, 3, 3√3,… must be taken to make the sum 39+ 13√3 ?
Solution: Given that, The sum of GP $=39+13 \sqrt{3}$ Where, $a=\sqrt{3}, r=3 / \sqrt{3}=\sqrt{3}, n=?$ Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ $39+13...
How many terms of the series 2 + 6 + 18 + …. Must be taken to make the sum equal to 728?
Solution: Given that, The sum of GP $=728$ Where, $a=2, r=6 / 2=3, n=?$ Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ $\begin{array}{l} 728=2\left(3^{n}-1\right)...
How many terms of the G.P. 3, 3/2, ¾, … Be taken together to make 3069/512 ?
Solution: Given that, The sum of G.P $=3069 / 512$ Where, $a=3, r=(3 / 2) / 3=1 / 2, n=?$ Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ $\begin{array}{l} 3069 /...
Find the sum of the following series:
(i) 0.6 + 0.66 + 0.666 + …. to n terms.
Solution: (i) $0.6+0.66+0.666+\ldots$ to $n$ terms. Let,s take 6 as a common term therefore we obtain, $6(0.1+0.11+0.111+\ldots n$ terms $)$ Now multiplying and dividing by 9 we obtain, $6 /...
Find the sum of the following series:
(i) 9 + 99 + 999 + … to n terms.
(ii) 0.5 + 0.55 + 0.555 + …. to n terms
Solution: (i) $9+99+999+\ldots$ to $n$ terms. We can write the given terms as $\begin{array}{l} (10-1)+(100-1)+(1000-1)+\ldots+n \text { terms } \\ \left(10+10^{2}+10^{3}+\ldots n \text { terms...
Find the sum of the following series:
(i) 5 + 55 + 555 + … to n terms.
(ii) 7 + 77 + 777 + … to n terms.
Solution: (i) $5+55+555+\ldots$ to $n$ terms. Let's take 5 as a common term therefore we obtain, $5[1+11+111+\ldots \mathrm{n}$ terms $]$ Now multiplying and dividing by 9 we obtain, $5 /...
Evaluate the following:
(i) $\sum_{n=2}^{10} 4^{n}$
Solution: (i) $\sum_{n=2}^{10} 4^{n}$ $=4^{2}+4^{3}+4^{4}+\ldots+4^{10}$ Where, $a=4^{2}=16, r=4^{3} / 4^{2}=4, n=9$ Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$...
Evaluate the following:
(i) $\sum_{n=1}^{11}\left(2+3^{n}\right)$
(ii) $\sum_{k=1 \atop 10}\left(2^{k}+3^{k-1}\right)$
Solution: (i) $\sum_{n=1}^{11}\left(2+3^{n}\right)$ $\begin{array}{l} =\left(2+3^{1}\right)+\left(2+3^{2}\right)+\left(2+3^{3}\right)+\ldots+\left(2+3^{11}\right) \\ =2 \times...
Find the sum of the following geometric series:
(i) $3 / 5+4 / 5^{2}+3 / 5^{3}+4 / 5^{4}+\ldots$ to $2 \mathrm{n}$ terms;
Solution: (i) $3 / 5+4 / 5^{2}+3 / 5^{3}+4 / 5^{4}+\ldots$ to $2 \mathrm{n}$ terms; We can write the series as: $3\left(1 / 5+1 / 5^{3}+1 / 5^{5}+\ldots\right.$ to $n$ terms $)+4\left(1 / 5^{2}+1 /...
Find the sum of the following geometric series:
(i) $2 / 9-1 / 3+1 / 2-3 / 4+\ldots$ to 5 terms;
(ii) $(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots$ to $n$ terms
Solution: (i) $2 / 9-1 / 3+1 / 2-3 / 4+\ldots$ to 5 terms; Given that $\begin{array}{l} a=2 / 9 \\ r=t_{2} / t_{1}=(-1 / 3) /(2 / 9)=-3 / 2 \\ n=5 \end{array}$ Using the formula, The sum of GP for...
Find the sum of the following geometric series:
(i) $0.15+0.015+0.0015+\ldots$ to 8 terms;
(ii) $\sqrt{2}+1 / \sqrt{2}+1 / 2 \sqrt{2}+\ldots$ to 8 terms;
Solution: (i) $0.15+0.015+0.0015+\ldots$ to 8 terms Given that $\begin{array}{l} a=0.15 \\ r=t_{2} / t_{1}=0.015 / 0.15=0.1=1 / 10 \\ n=8 \end{array}$ Using the formula, The sum of GP for $n$ terms...
Find the sum of the following geometric progressions:
(i) $4,2,1,1 / 2 \ldots$ to 10 terms
Solution: (i) $4,2,1,1 / 2 \ldots$ to 10 terms It is known that, the sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ Given that, $\mathrm{a}=4, \mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=2 /...
Find the sum of the following geometric progressions:
(i) $1,-1 / 2,1 / 4,-1 / 8, \ldots$
(ii) $\left(a^{2}-b^{2}\right),(a-b),(a-b) /(a+b), \ldots$ to $n$
Solution: (i) $1,-1 / 2,1 / 4,-1 / 8, \ldots$ It is known that, the sum of $\mathrm{GP}$ for infinity $=\mathrm{a} /(1-\mathrm{r})$ Given that, $\mathrm{a}=1, \mathrm{r}=\mathrm{t}_{2} /...
Find the sum of the following geometric progressions:
(i) 2, 6, 18, … to 7 terms
(ii) 1, 3, 9, 27, … to 8 terms
Solution: (i) $2,6,18, \ldots$ to 7 terms It is known that, the sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ Given that, $a=2, r=t_{2} / t_{1}=6 / 2=3, n=7$ Substitute the values in...
The sum of the first three terms of a G.P. is 39 / 10, and their product is 1 . Find the common ratio and the terms.
Solution: Let the three numbers be $a / r, a, ar$ According to the question $\mathrm{a} / \mathrm{r}+\mathrm{a}+\mathrm{ar}=39 / 10 \ldots$ eq. (1) $\mathrm{a} / \mathrm{r} \times \mathrm{a} \times...
The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is 871 / 2. Find them.
Solution: Let the three numbers be $a/r$, $a$, $ar$ According to the question $\mathrm{a} / \mathrm{r} \times \mathrm{a} \times \mathrm{ar}=125 \ldots$ eq.(1) From eq.(1) we obtain,...
The sum of first three terms of a G.P. is 13 / 12, and their product is -1 . Find the G.P.
Solution: Let the three numbers be $a/r$, $a$, $ar$ According to the question $\mathrm{a} / \mathrm{r}+\mathrm{a}+\mathrm{ar}=13 / 12 \ldots$ eq. (1) $\mathrm{a} / \mathrm{r} \times \mathrm{a}...
Find three number in G.P. whose sum is 38 and their product is 1728.
Solution: Let the three numbers be $\mathrm{a} / \mathrm{r}, \mathrm{a}$, $ar$ According to the question, $\mathrm{a} / \mathrm{r}+\mathrm{a}+\mathrm{ar}=38 \ldots$ eq.(1) $\mathrm{a} / \mathrm{r}...
Find three numbers in G.P. whose sum is 65 and whose product is 3375.
Solution: According to the question $\mathrm{a} / \mathrm{r}+\mathrm{a}+\mathrm{ar}=65 \ldots$ equation (1) $a / r \times a \times a r=3375 \ldots$ equation (2) From eq.(2) we obtain,...
Find the 4th term from the end of the G.P. ½, 1/6, 1/18, 1/54, … , 1/4374
Solution: $n^{th}$ term from the end is given by: $a_{n}=I(1 / r)^{n-1}$ where, $I$ is the last term, $r$ is the common ratio, $n$ is the $n^{th}$ term Given that, last term, $I=1 / 4374$...
Which term of the progression 18, -12, 8, … is 512/729 ?
Solution: Using the formula, $\begin{array}{l} \mathrm{T}_{n}=\mathrm{ar}^{\mathrm{n}-1} \\ \mathrm{a}=18 \\ \mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=(-12 / 18) \\ =-2 / 3 \\...
Which term of the G.P.:
(i) √3, 3, 3√3, … is 729 ?
(ii) 1/3, 1/9, 1/27… is 1/19683 ?
Solution: (i) $\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is $729 ?$ Using the formula, $\begin{array}{l} \mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1} \\ \mathrm{a}=\sqrt{3} \\ \mathrm{r}=\mathrm{t}_{2}...
Which term of the G.P.:
(i) √2, 1/√2, 1/2√2, 1/4√2, … is 1/512√2 ?
(ii) 2, 2√2, 4, … is 128 ?
Solution: (i) $\sqrt{2}, 1 / \sqrt{2}, 1 / 2 \sqrt{2}, 1 / 4 \sqrt{2}, \ldots$ is $1 / 512 \sqrt{2} ?$ Using the formula, $\begin{array}{l} T_{n}=a r^{n-1} \\ a=\sqrt{2} \\ r=t_{2} / t_{1}=(1 /...
Which term of the progression 0.004, 0.02, 0.1, …. is 12.5?
Solution: Using the formula, $\mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}$ Given that, $\begin{array}{l} a=0.004 \\ r=t_{2} / t_{1}=(0.02 / 0.004) \\ =5 \\ T_{n}=12.5 \\ n=? \end{array}$...
Find the $4^{\text {th }}$ term from the end of the G.P. $2 / 27,2 / 9,2 / 3, \ldots ., 162$.
Solution: The $n^{th}$ term from the end is given by: $a_{n}=I(1 / r)^{n-1}$ where, $I$ is the last term, $r$ is the common ratio, $n$ is the nth term Given that, last term, $\mid=162$...
Find:
(i) $n^{th}$ term of the G.P. $\sqrt{3}, 1 / \sqrt{3}, 1 / 3 \sqrt{3}, \ldots$
(ii) the $10^{\text {th }}$ term of the G.P. $\sqrt{2}, 1 / \sqrt{2}, 1 / 2 \sqrt{2}, \ldots .$
Solution: (i) nth term of the G.P. $\sqrt{3}, 1 / \sqrt{3}, 1 / 3 \sqrt{3}, \ldots$ It is known that, $t_{1}=a=\sqrt{3}, r=t_{2} / t_{1}=(1 / \sqrt{3}) / \sqrt{3}=1 /(\sqrt{3} \times \sqrt{3})=1 /...
Find:
(i) the $8^{\text {th }}$ term of the G.P. $0.3,0.06,0.012, \ldots .$
(ii) the $12^{\text {th }}$ term of the G.P. $1 / a^{3} x^{3}, a x, a^{5} x^{5}, \ldots .$
Solution: (i) the $8^{\text {th }}$ term of the G.P., $0.3,0.06,0.012, \ldots$ It is known that, $\mathrm{t}_{1}=\mathrm{a}=0.3, \mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=0.06 / 0.3=0.2$ Using the...
Find:
(i) the ninth term of the G.P. $1,4,16,64, \ldots .$
(ii) the $10^{\text {th }}$ term of the G.P. $-3 / 4,1 / 2,-1 / 3,2 / 9, \ldots .$
Solution: (i) the ninth term of the G.P. $1,4,16,64, \ldots$ It is known that, $t_{1}=a=1, r=t_{2} / t_{1}=4 / 1=4$ Using the formula. $\begin{array}{l}...
Show that the sequence defined by $a_{n}=2 / 3^{n}, n \in N$ is a G.P.
Solution: Given that, $a_{n}=2 / 3^{n}$ Consider $\mathrm{n}=1,2,3,4, \ldots$ since $\mathrm{n}$ is a natural number. Therefore, $\begin{array}{l} a_{1}=2 / 3 \\ a_{2}=2 / 3^{2}=2 / 9 \\ a_{3}=2 /...
Show that each one of the following progressions is a G.P. Also, find the common ratio in each case:
(i) $a, 3 a^{2} / 4,9 a^{3} / 16, \ldots$
(ii) $1 / 2,1 / 3,2 / 9,4 / 27, \ldots$
Solution: (i) a, $3 \mathrm{a}^{2} / 4,9 \mathrm{a}^{3} / 16, \ldots$ Let $a=a, b=3 a^{2} / 4, c=9 a^{3} / 16$ In Geometric Progression, $\begin{array}{l} b^{2}=a c \\ \left(3 a^{2} / 4\right)^{2}=9...
Show that each one of the following progressions is a G.P. Also, find the common ratio in each case:
(i) $4,-2,1,-1 / 2, \ldots$
(ii) $-2 / 3,-6,-54, \ldots .$
Solution: (i) $4,-2,1,-1 / 2, \ldots$ Let $a=4, b=-2, c=1$ In $\mathrm{GP}$ $\begin{array}{l} b^{2}=a c \\ (-2)^{2}=4(1) \\ 4=4 \end{array}$ Therefore, the Common ratio $=r=-2 / 4=-1 / 2$ (ii) $-2 /...
Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.
Solution: Let a and b be the roots of the quadratic equation. So, in accordance to the given condition, we have $ A.M\text{ }=\text{ }\left( a+b \right)/2\text{ }=\text{ }A $ $ a\text{ }+\text{...
Find the two numbers whose A.M. is 25 and GM is 20.
Solution: It is given that A.M = 25, G.M = 20. We have, G.M = √ab and A.M = (a+b)/2 So, we can write: √ab = 20 ……. (1) (a+b)/2 = 25……. (2) Also, a + b = 50 Or, a = 50 – b Substituting the value of...
If a is the G.M. of 2 and ¼ find a.
Solution: According to the question, the GM between a and b is √ab We have a = 2 and b = 1/4 Therefore, GM = √(2×1/4)= √(1/2) GM = 1/√2 Therefore, the value of a is 1/√2
Find the geometric means of the following pairs of numbers:
(iii) –8 and –2 Solution: (iii) –8 and –2 According to the question, GM between a and b is √ab we have, a = –2 and b = –8 GM = √(–2×–8) = √–16 GM = -4
Find the geometric means of the following pairs of numbers:
(i) 2 and 8 (ii) a3b and ab3 Solution: (i) 2 and 8 According to the question, GM between a and b is √ab We have a = 2 and b =8 GM = √2×8 = √16 GM = 4 (ii) a3b and ab3 According to the question, GM...
Insert 5 geometric means between 32/9 and 81/2.
Solution: Suppose that the five terms are a1, a2, a3, a4, a5. where A = 32/9, B = 81/2 The above five terms are between A and B. So the GP is: A, a1, a2, a3, a4, a5, B. Therefore, there are 7 terms...
Insert 5 geometric means between 16 and 1/4.
Solution: Consider the first five terms to be a1, a2, a3, a4, a5. we have, A = 27, B = 1/81 The above mentioned five terms are between A and B. So the GP is: A, a1, a2, a3, a4, a5, B. Therefore, we...
Insert 6 geometric means between 27 and 1/81.
Solution: Let a1, a2, a3, a4, a5, a6 be the six terms We have, A = 27, B = 1/81 According to the question, these 6 terms are between A and B. So the GP is given by: A, a1, a2, a3, a4, a5, a6, B. So...
If a, b, c are in G.P., prove that the following are also in G.P.:
(iii) a2 + b2, ab + bc, b2 + c2 Solution: (iii) a2 + b2, ab + bc, b2 + c2 According to the question, a, b, c are in GP. Making use of the property of geometric mean, b2 = ac a2 + b2, ab + bc, b2 +...
If a, b, c are in G.P., prove that the following are also in G.P.:
(i) a2, b2, c2 (ii) a3, b3, c3 Solution: (i) a2, b2, c2 According to the question, a, b, c are in GP. Making use of the property of geometric mean, we can write: b2 = ac Upon squaring both the sides...
If a, b, c, d are in G.P., prove that:
(iii) (b + c) (b + d) = (c + a) (c + d) Solution: (iii) (b + c) (b + d) = (c + a) (c + d) According to the question, a, b, c are in GP. Making use of the property of geometric mean, we can write:...
If a, b, c, d are in G.P., prove that:
(i) (ab – cd) / (b2 – c2) = (a + c) / b (ii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2 Solution: (i) (ab – cd) / (b2 – c2) = (a + c) / b According to the question a, b, c are in GP. Making...
if a, b, c are in G.P., prove that:
(v) (a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2 Solution: (v) (a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2 According to the question, a, b, c are in GP. Making use of the property of geometric mean, b2 = ac...
if a, b, c are in G.P., prove that:
(iii) (a+b+c)2 / (a2 + b2 + c2) = (a+b+c) / (a-b+c) (iv) 1/(a2 – b2) + 1/b2 = 1/(b2 – c2) Solution: (iii) (a+b+c)2 / (a2 + b2 + c2) = (a+b+c) / (a-b+c) According to the question, a, b, c are in GP....
if a, b, c are in G.P., prove that:
(i) a(b2 + c2) = c(a2 + b2) (ii) a2b2c2 [1/a3 + 1/b3 + 1/c3] = a3 + b3 + c3 Solution: (i) a(b2 + c2) = c(a2 + b2) According to the question, a, b, c are in GP. Making use of the property of...
The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.
Solution: Suppose that the three numbers are a, ar, ar2 According to the question statement, a + ar + ar2 = 56 … (1) subtracting 1,7,21 respectively from these terms,we get: (a – 1), (ar – 7),...
The sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36, the new numbers form a G.P. Find a, b, c.
Solution: Consider the first term of an A.P. to be ‘a’ and let its common difference be given by ‘d’. Therefore, we can write: b = a + d and c = a + 2d. It is gien that: a + b + c = 18 Or, we can...
The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.
Solution: Consider the first term of an A.P. to be ‘a’ and let its common difference be‘d’. Therefore, we can write: a1 + a2 + a3 = 21 Where the three numbers are as follows: a, a + d, and a + 2d...
Three numbers are in A.P., and their sum is 15. If 1, 3, 9 be added to them respectively, they from a G.P. find the numbers.
Solution: Consider the first term of an A.P. to be ‘a’ and let its common difference be‘d’. We have, a1 + a2 + a3 = 15 Where the three numbers are as follows: a, a + d, and a + 2d So, we can write:...
Find k such that k + 9, k – 6 and 4 form three consecutive terms of a G.P.
Solution: Suppose that a = k + 9; b = k−6; and c = 4; According to the question, a, b and c are in GP, then Making use of the property of geometric mean, we get: b2 = ac $ {{\left( k-6...
If a, b, c are in G.P., prove that 1/loga m , 1/logb m, 1/logc m are in A.P.
Solution: According to the question, a, b and c are in GP Making use of the property of geometric mean, we can write: b2 = ac Taking log on both sides with base m, we get: $...
If a, b, c are in G.P., prove that log a, log b, log c are in A.P.
Solution: According to the question, a, b and c are in G.P. By making use of the property of Geometric mean, we can write: $ {{b}^{2}}=~ac $ $ ~{{\left( {{b}^{2}} \right)}^{n}}~=~{{\left( ac...
Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is equal to 32/81.
Solution: Let ‘a’ be the first term of GP and ‘r’ be the common ratio. We know that nth term of a GP is given by- an = arn-1 As, a = 4 (given) And a5 – a3 = 32/81 (given) 4r4 – 4r2 = 32/81 4r2(r2 –...
Express the recurring decimal 0.125125125 … as a rational number.
Solution: We can write 0.125125125 as 0.125125125 = 0.125 + 0.000125 + 0.000000125 + … This can be written as $ 125/{{10}^{3}}~+\text{ }125/{{10}^{6}}~+\text{ }125/{{10}^{9}}~+\text{ }\ldots $...
If Sp denotes the sum of the series 1 + rp + r2p + … to ∞ and sp the sum of the series 1 – rp + r2p – … to ∞, prove that sp + Sp = 2 S2p.
Solution: Given: Sp = 1 + rp + r2p + … ∞ By using the formula, S∞ = a/(1 – r) Where, a = 1, r = rp So, Sp = 1 / (1 – rp) Similarly, sp = 1 – rp + r2p – … ∞ By using the formula, S∞ = a/(1 – r)...
Prove that : \[({{\mathbf{2}}^{\mathbf{1}/\mathbf{4}}}~.{{\mathbf{4}}^{\mathbf{1}/\mathbf{8}}}~.\text{ }{{\mathbf{8}}^{\mathbf{1}/\mathbf{16}}}.\text{ }\mathbf{1}{{\mathbf{6}}^{\mathbf{1}/\mathbf{32}}}\ldots .\infty )\text{ }=\text{ }\mathbf{2}\]
Solution: We can write the LHS of above equation as: ${{2}^{1/4}}~.\text{ }{{2}^{2/8}}~.\text{ }{{2}^{3/16}}~.\text{ }{{2}^{1/8}}~\ldots \text{ }\infty $ ${{2}^{1/4\text{ }+\text{ }2/8\text{...
Prove that : \[\left( {{\mathbf{9}}^{\mathbf{1}/\mathbf{3}}}~.\text{ }{{\mathbf{9}}^{\mathbf{1}/\mathbf{9}}}~.\text{ }{{\mathbf{9}}^{\mathbf{1}/\mathbf{27}}}~\ldots .\infty \right)\text{ }=\text{ }\mathbf{3}.\]
Solution: Let us take the LHS first: We can write the given equation as: 91/3 + 1/9 + 1/27 + …∞ So let us take $ m\text{ }=\text{ }1/3\text{ }+\text{ }1/9\text{ }+\text{ }1/27\text{ }+\text{ }\ldots...
Find the sum of the following series to infinity:
(iii) 2/5 + 3/52 + 2/53 + 3/54 + …. ∞ (iv) 10 – 9 + 8.1 – 7.29 + …. ∞ Solution: (iii) 2/5 + 3/52 + 2/53 + 3/54 + …. ∞ We can write the given terms as, (2/5 + 2/53 + …) + (3/52 + 3/54 + …) where (a =...
Find the sum of the following series to infinity:
(i) 1 – 1/3 + 1/32 – 1/33 + 1/34 + … ∞ (ii) 8 + 4√2 + 4 + …. ∞ Solution: (i) 1 – 1/3 + 1/32 – 1/33 + 1/34 + … ∞ We are given that: S∞ = 1 – 1/3 + 1/32 – 1/33 + 1/34 + … ∞ On comparing with the...