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Visual SolutionPYQ 2024 · Jan 29 Shift 1Easy
Question
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)77
(B)44
(C)55
(D)66
Solution Path
GP sum formula → odd terms form GP with ratio r2r^2 → ratio cancels to r+1=7r + 1 = 7 → r=6r = 6
01Read the Problem
1/5
GP with 64 terms. The total sum is 7 times the sum of odd-positioned terms. We need the common ratio rr.
S=7×SoddS = 7 \times S_{\text{odd}}
02Set Up the GP
2/5
Standard GP sum formula with first term aa, common ratio rr, and 64 terms.
S=a⋅r64−1r−1S = a \cdot \dfrac{r^{64} - 1}{r - 1}
03Odd-Positioned Terms
3/5
The odd-positioned terms a,ar2,ar4,…,ar62a, ar^2, ar^4, \ldots, ar^{62} form their own GP with common ratio r2r^2 and 32 terms.
Sodd=a⋅r64−1r2−1S_{\text{odd}} = a \cdot \dfrac{r^{64} - 1}{r^2 - 1}
04The Elegant CancellationKEY INSIGHT
4/5
When you divide SS by SoddS_{\text{odd}}, both aa and (r64−1)(r^{64}-1) cancel completely, leaving just (r2−1)/(r−1)=r+1(r^2-1)/(r-1) = r+1.
SSodd=r+1\dfrac{S}{S_{\text{odd}}} = r + 1
05Final Answer
5/5
Since S/Sodd=7S/S_{\text{odd}} = 7, we get r+1=7r + 1 = 7.
r=6r = \boxed{6}
Concepts from this question4 concepts unlocked
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GP Subsequence Property

STANDARD

Every equally-spaced subsequence of a GP is itself a GP with a power of the original ratio

Every k-th term: ratio =rk,terms=⌈n/k⌉\text{Every } k\text{-th term: ratio } = r^k, \quad \text{terms} = \lceil n/k \rceil

Odd/even positioned terms, every 3rd term, etc. all form GPs. This pattern appears in 3-4 questions per year.

Odd/even term sumsAlternating seriesSubsequence ratios
Practice (6 Qs) →
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Ratio Cancellation in GP Sums

STANDARD

When dividing two GP sums with the same base, a and r^n terms cancel, leaving a simple expression in r

SSodd=r2−1r−1=r+1\frac{S}{S_{\text{odd}}} = \frac{r^2 - 1}{r - 1} = r + 1

The 'messy' parts (a, r^n) always cancel. Recognizing this saves 3-4 minutes of algebra.

Sum ratio problemsCondition-based GPInteger ratio constraints
Practice (7 Qs) →

GP Sum Formula

EASY

Sum of n terms of a GP with first term a and common ratio r

Sn=a⋅rn−1r−1(r≠1)S_n = a \cdot \frac{r^n - 1}{r - 1} \quad (r \neq 1)

Foundation for nearly every GP problem in JEE. Must be instant recall.

Sum problemsRatio conditionsInfinite GP
Practice (12 Qs) →

Difference of Squares Factoring

EASY

r^2 - 1 = (r-1)(r+1) allows cancellation with (r-1) denominators

r2−1r−1=(r−1)(r+1)r−1=r+1\frac{r^2 - 1}{r - 1} = \frac{(r-1)(r+1)}{r-1} = r + 1

This algebraic identity is the final key step in many GP ratio problems. Without it, you get stuck.

GP simplificationPolynomial factoringSeries telescoping
Practice (9 Qs) →
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