JEE Main 2022 · 28 June · morning · Single correct · +4 / −1
Let A₁, A₂, A₃, ....... be an increasing geometric progression of positive real numbers. If A₁A₃A₅A₇ = 1/1296 and A₂ + A₄ = 7/36, then the value of A₆ + A₈ + A₁₀ is equal to
- (a) 33
- (b) 37
- (c) 43
- (d) 47
Answer and solution
Answer (c) 43
Solution
Maths sir · 4 steps
Pair the terms from the ends: A₁A₇ and A₃A₅ both equal A₄², so the product hands you A₄.
- Pair the terms. In a GP, terms equally far from a middle term multiply to that term squared. So A₁A₇ = A₄² and A₃A₅ = A₄², which makes the given product A₄⁴.
A₄⁴ = 1/1296 = 1/6⁴ ⇒ A₄ = 1/6
- Find the ratio. All terms are positive, so A₄ = 1/6 = 6/36. Then A₂ = 7/36 - 6/36 = 1/36, and r² = A₄/A₂ = 6. This is greater than 1, which fits an increasing GP.
- Climb in steps. Each even term is r² = 6 times the one before it. So A₆ = 1/6 × 6 = 1, A₈ = 6 and A₁₀ = 36.
- So. A₆ + A₈ + A₁₀ = 1 + 6 + 36 = **43**, which is option (c).
The idea to keep. In a GP, the product of terms equally spaced around a centre is a power of the centre term. When only even-numbered terms appear, work with r² and you never need r itself.
Answer from the NTA final answer key. Solution by PrepWiser.