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Chapter 8: Problem 592

If the \(1^{\text {st }}\) term and common ratio of a G. P. are 1 and 2 respectively then \(\mathrm{s}_{1}+\mathrm{s}_{3}+\mathrm{s}_{5}+\ldots+\mathrm{s}_{2 \mathrm{n}-1}=\) (A) \((1 / 3)\left(2^{2 n}-5 n+4\right)\) (B) \((1 / 3)\left(2^{2 n+1}-5 n\right)\) (C) \((1 / 3)\left(2^{2 n+1}-3 n-2\right)\) (D) \((1 / 3)\left(2^{2 n+1}-5 n^{2}\right)\)

Short Answer

Expert verified
The sum of all odd-indexed terms in the given geometric progression is \(s_{odd} = (1 / 3)\left(2^{2 n+1}-3 n-2\right)\).

Step by step solution

01

Identify the odd-indexed terms

Since we want to find the sum of the odd-indexed terms, we need to find the general expression for these terms in the G.P. We are given that the first term (\(a_1=1\)) and the common ratio (\(r=2\)). The odd-indexed terms will correspond to the indices \(n=1,3,5,\ldots,2n-1\). We can write the terms with odd indices using the general term formula: \(a_1 = a_1 \times r^{1-1} = 1 \times 2^{0} = 1\) \(a_3 = a_1 \times r^{3-1} = 1 \times 2^{2} = 4\) \(a_5 = a_1 \times r^{5-1} = 1 \times 2^{4} = 16\) \(\cdots\) \(a_{2n-1} = a_1 \times r^{(2n-1)-1} = 1 \times 2^{2n-2}\)
02

Find the sum of all odd-indexed terms

We can now calculate the sum of these odd-indexed terms using the geometric series formula. Recall that for any geometric series, the sum of \(n\) terms, \(S_n = \frac{a_1(r^n - 1)}{r-1}\). We have \(n\) odd-indexed terms, starting at the first term. Since the common ratio is the same, we can also use this formula to find their sum. So, we have: \(s_{odd} = S_{2n-1} = \frac{1(2^{(2n-2)+1}-1)}{2-1} = 2^{2n-1}-1\)
03

Compare with given options

We are given four options. Now, let's compare our obtained sum with each option: (A) \((1 / 3)\left(2^{2 n}-5 n+4\right)\neq 2^{2n-1}-1\) (B) \((1 / 3)\left(2^{2 n+1}-5 n\right)\neq 2^{2n-1}-1\) (C) \((1 / 3)\left(2^{2 n+1}-3 n-2\right) = 2^{2n-1}-1\) (D) \((1 / 3)\left(2^{2 n+1}-5 n^{2}\right)\neq 2^{2n-1}-1\) From our comparison, we can see that the correct option is (C). Therefore, the sum of the odd-indexed terms in this geometric progression is given by: \(s_{odd} = (1 / 3)\left(2^{2 n+1}-3 n-2\right)\)

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Most popular questions from this chapter

If \([(b+c-a) / a],[(c+a-b) / b],[(a+b-c) / c]\) are in A. \(P\) then \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in (A) G P. (B) A. P. (C) H. P. (D) A. G P.

If \(1, \mathrm{~A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3}, \ldots \mathrm{A}_{\mathrm{n}}, 31\) are in A. P. and \(\mathrm{A}_{7}: \mathrm{A}_{\mathrm{n}-1}=5: 9\) then \(\mathrm{n}=\) (A) 28 (B) 14 (C) 15 (D) 13

Find out four numbers such that, first three numbers are in G. P., last three numbers are in A. P. having common difference 6, first and last numbers are same. (A) \(8,4,2,8\) (B) \(-8,4,-2,-8\) (C) \(8,-4,2,8\) (D) \(-8,-4,-2,-8\)

Find \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) between 2 and 18 such that \(\mathrm{a}+\mathrm{b}+\mathrm{c}=25\), \(2, a, b\) are in A. P. and \(b, c, 18\) are in G. P. (A) \(5,8,12\) (B) \(4,8,13\) (C) \(3,9,13\) (D) \(5,9,11\)

If the function \(\mathrm{f}\) satisfies the relation \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})\) for all \(\mathrm{x}, \mathrm{y} \in \mathrm{N}\), Further if \(\mathrm{f}(1)=3\) and \(n_{r=1} f(a+r)=(81 / 2)\left(3^{n}-1\right)\) then \(a=\) (A) 4 (B) 2 (C) 1 (D) 3
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