Suppose a₁, a₂, 2, a₃, a₄ be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is 49/2, then a₄ is equal to __________.

Solution

Maths sir · 4 steps

Build every term around the middle term 2; then the sum gives one equation in d and you are done.

1. General term. In an AGP the nth term is (AP term) × (GP term), so Tₙ = (a + (n-1)d) 2ⁿ⁻¹. The middle term is T₃ = (a+2d) × 4 = 2, so a + 2d = 1/2.
2. Centre the AP. Let b = a + 2d = 1/2. The five AP parts are b-2d, b-d, b, b+d, b+2d, multiplied by 1, 2, 4, 8, 16.

   S = b(1+2+4+8+16) + d(-2-2+0+8+32) = 31b + 36d

3. Find d. 31 × 1/2 + 36d = 49/2, so 36d = 9 and d = 1/4. The terms are 0, 1/2, 2, 6, 16, and they do add to 49/2.
4. So. a₄ is the fifth term: (b+2d) × 16 = (1/2+1/2) × 16 = 16.

The common slip: 6. Treating a₄ as the fourth term of the progression gives 6. Here the labels skip the known term 2, so a₄ is the fifth term.

The idea to keep. In an AGP, centre the AP part on the term you know. One known term plus one sum gives two equations, and the unknowns fall out quickly.

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