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Visual SolutionPYQ 2024 · Jan (1 Feb) Shift 1Standard
Question
Let 3,a,b,c3, a, b, c be in A.P. and 3,a−1,b+1,c+93, a-1, b+1, c+9 be in G.P. Then, the arithmetic mean of aa, bb and cc is:
(A)−4-4
(B)−1-1
(C)1313
(D)1111
Solution Path
AP with dd gives a=3+da=3+d, b=3+2db=3+2d, c=3+3dc=3+3d. GP condition (2+d)2=3(4+2d)(2+d)^2 = 3(4+2d) gives d=4d=4. AM =(7+11+15)/3=11= (7+11+15)/3 = 11.
01Question Setup
1/4
3,a,b,c3, a, b, c in A.P. and 3,a−1,b+1,c+93, a-1, b+1, c+9 in G.P. Find the arithmetic mean of a,b,ca, b, c.
Find AM(a,b,c)\text{AM}(a, b, c)
02A.P. Setup
2/4
With common difference dd: a=3+da = 3+d, b=3+2db = 3+2d, c=3+3dc = 3+3d. The G.P. becomes 3,2+d,4+2d,12+3d3, 2+d, 4+2d, 12+3d.
a=3+d,  b=3+2d,  c=3+3da = 3+d, \; b = 3+2d, \; c = 3+3d
03G.P. ConditionKEY INSIGHT
3/4
In a G.P., (2nd)2=(1st)(3rd)(2\text{nd})^2 = (1\text{st})(3\text{rd}): (2+d)2=3(4+2d)(2+d)^2 = 3(4+2d). Simplifies to d2−2d−8=0d^2 - 2d - 8 = 0, giving d=4d = 4 (reject d=−2d = -2).
d2−2d−8=0  ⟹  d=4d^2 - 2d - 8 = 0 \implies d = 4
04Final Answer
4/4
With d=4d = 4: a=7,b=11,c=15a = 7, b = 11, c = 15. AM =(7+11+15)/3=11= (7+11+15)/3 = 11.
AM=333=11\text{AM} = \dfrac{33}{3} = \boxed{11} - Answer (D)
Concepts from this question2 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) →

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) →
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