x = Σₙ₌₀^∞ aⁿ,y = Σₙ₌₀^∞ bⁿ,z = Σₙ₌₀^∞ cⁿ , where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1, abc ≠ 0, then :

Solution

Maths sir · 3 steps

Sum each infinite GP first, then flip it: 1/x = 1 - a is linear in a.

1. Sum each series. Each series is an infinite GP with first term 1 and ratio less than 1 in size, so it converges. For example, x = 1 + a + a² + … = 1/(1-a).

   1/x = 1 - a, 1/y = 1 - b, 1/z = 1 - c

2. Use the A.P. Since a, b, c are in A.P., 2b = a + c. Then 2(1 - b) = 2 - a - c = (1 - a) + (1 - c), so 2/y = 1/x + 1/z.
3. So. 1/x, 1/y, 1/z are in A.P., which is option (c). Equivalently, x, y, z are in H.P.

The idea to keep. Subtracting every term of an A.P. from the same constant still gives an A.P. Reciprocals of A.P. terms form an H.P., not an A.P.

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