Let a₁, a₂, ......., a₁₀ be an AP with common difference - 3 and b₁, b₂, ........., b₁₀ be a GP with common ratio 2. Let cₖ = aₖ + bₖ, k = 1, 2, ......, 10. If c₂ = 12 and c₃ = 13, then Σₖ₌₁¹⁰ cₖ is equal to _________.

Solution

Maths sir · 4 steps

Write c₂ and c₃ in terms of a₁ and b₁, solve the pair, then add the AP and GP sums separately.

1. Set up equations. With aₖ = a₁ - 3(k-1) and bₖ = b₁ · 2ᵏ⁻¹, we get c₂ = a₁ - 3 + 2b₁ = 12 and c₃ = a₁ - 6 + 4b₁ = 13.
2. Solve the pair. Subtract the first equation from the second: -3 + 2b₁ = 1, so b₁ = 2. Then a₁ - 3 + 4 = 12 gives a₁ = 11.

   Σₖ₌₁¹⁰ cₖ = 10/2[2(11) + 9(-3)] + 2(2¹⁰-1)/(2-1)

3. Add the parts. The AP part is 5 × (22 - 27) = -25. The GP part is 2 × 1023 = 2046.
4. So. Σₖ₌₁¹⁰ cₖ = -25 + 2046 = 2021.

The common slip: 998. Forgetting to multiply the GP sum by b₁ = 2 gives 1023 - 25 = 998, not 2021.

The idea to keep. When a sequence is AP plus GP, two given terms fix the two unknown first terms. Then sum the AP and the GP on their own and add.

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