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Sequences and Series question

2024 · 29 Jan · Shift 1 · Q35

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2024 · 29 Jan · Shift 1 · Q35

Sequences and Series question

2024 · 29 Jan · Shift 1 · Q35

JEE MainMathematicsSequences and SeriesMCQ+4 / −1

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

A
7
B
6
C
5
D
4

View written solutionFree

Correct answer: B

Let the G.P. have first term $a$ and common ratio $r$ .
Sum of all $64$ terms:

S_{64}=a\frac{1-r^{64}}{1-r}\quad (r\ne 1)

The odd-positioned terms are:

a,\ ar^2,\ ar^4,\ \dots,\ ar^{62}

This is itself a G.P. with:

first term $a$
common ratio $r^2$
number of terms $32$

So, sum of odd terms is

S_{\text{odd}}=a\frac{1-(r^2)^{32}}{1-r^2}=a\frac{1-r^{64}}{1-r^2}

Given:

S_{64}=7S_{\text{odd}}

Substitute the formulas:

a\frac{1-r^{64}}{1-r}=7\left(a\frac{1-r^{64}}{1-r^2}\right)

Assuming the G.P. is nontrivial, cancel $a(1-r^{64})$ from both sides:

\frac{1}{1-r}=\frac{7}{1-r^2}

Since

1-r^2=(1-r)(1+r),

we get

\frac{1}{1-r}=\frac{7}{(1-r)(1+r)}

Cancelling $(1-r)$ :

1=\frac{7}{1+r}

So,

1+r=7

r=6

Check with options:

A: $7$
B: $6$ ✅
C: $5$
D: $4$

Therefore, the common ratio is

\boxed{6}

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