The sum of first four terms of a geometric progression (G. P.) is 65/12 and the sum of their respective reciprocals is 65/18. If the product of first three terms of the G.P. is 1, and the third term is α, then 2α is _________.

Solution

Maths sir · 5 steps

Divide the sum by the sum of reciprocals: the long bracket cancels and leaves a neat a²r³.

1. Set up. Take the terms as a, ar, ar², ar³. Their sum is a(1+r+r²+r³) = 65/12. The reciprocals sum to 1/a(1+1/r+1/r²+1/r³) = (1+r+r²+r³)/ar³ = 65/18.
2. Divide the two. The bracket 1+r+r²+r³ is the same in both, so it cancels when you divide.

   a(1+r+r²+r³)/((1+r+r²+r³)/ar³) = a²r³ = (65/12)/(65/18) = 3/2

3. Use the product. The product of the first three terms is a · ar · ar² = (ar)³ = 1, so ar = 1. Then a²r³ = (ar)² × r = r, which gives r = 3/2 and a = 2/3.
4. Check. The terms are 2/3, 1, 3/2, 9/4. Their sum is (8+12+18+27)/12 = 65/12, as required.
5. So. The third term is α = ar² = (ar) × r = 3/2, so 2α = 3.

The common slip: 1.5. Reporting α itself gives 1.5. The question asks for 2α, which is 3.

The idea to keep. For a GP, the sum of terms divided by the sum of their reciprocals kills the common bracket. If a product of terms is given, turn it into a single power like (ar)³ first.

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