If each term of a geometric progression a₁, a₂, a₃, … with a₁=1/8 and a₂ ≠ a₁, is the arithmetic mean of the next two terms and Sₙ=a₁+a₂+… . .+aₙ, then S₂₀-S₁₈ is equal to

Solution

Maths sir · 4 steps

Turn the mean condition into an equation in r, then S₂₀-S₁₈ is just the last two terms.

1. Find the ratio. Write aₙ = arⁿ⁻¹. The condition aₙ = (aₙ₊₁+aₙ₊₂)/2 becomes 2arⁿ⁻¹ = arⁿ + arⁿ⁺¹. Dividing by arⁿ⁻¹ gives r² + r - 2 = 0.
2. Reject r = 1. The roots are r = 1 and r = -2. Since a₂ ≠ a₁, r ≠ 1, so r = -2.
3. Two terms left. S₂₀ - S₁₈ = a₁₉ + a₂₀, because every earlier term cancels. Factor out a₁₉.

   a₁₉ + a₂₀ = a₁ r¹⁸(1 + r) = 1/8 × 2¹⁸ × (-1) = -2¹⁵

4. So. S₂₀ - S₁₈ = -2¹⁵, which is option (a).

The idea to keep. A difference of two partial sums is just the terms between them. Never compute the whole Sₙ when only a few terms survive.

Answer from the NTA final answer key. Solution by PrepWiser.