If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

Solution

Maths sir · 4 steps

Divide the two sums: everything cancels except a power of r, so r falls out in one line.

1. Write both sums. With first term a and ratio r, the second, fourth and sixth terms add to ar(1 + r² + r⁴) = 21. The eighth, tenth and twelfth terms add to ar⁷(1 + r² + r⁴) = 15309.
2. Find r. Divide the second sum by the first. The common factor ar(1 + r² + r⁴) cancels, leaving r⁶ = 15309/21 = 729 = 3⁶. All terms are positive, so r = 3 and not -3.
3. Find a. Put r = 3 into the first sum: 3a(1 + 9 + 81) = 273a = 21, so a = 21/273 = 1/13.

   S₉ = a(r⁹ - 1)/(r - 1) = 1/13 × (19683 - 1)/(3 - 1) = 19682/26 = 757

4. So. The sum of the first nine terms is 757, option (a).

The idea to keep. When two sums of a G.P. share the same pattern of terms, divide one by the other. The common bracket cancels and you are left with a clean power of r.

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