Suppose a₁, a₂, …, aₙ, .. be an arithmetic progression of natural numbers. If the ratio of the sum of first five terms to the sum of first nine terms of the progression is 5: 17 and , 110 < a₁₅ < 120, then the sum of the first ten terms of the progression is equal to

Solution

Maths sir · 3 steps

Turn the ratio into a link between a and d, and then the bounds on a₁₅ fix a.

1. Write both sums. Let the first term be a and the common difference be d. Then S₅ = 5/2(2a + 4d) = 5(a + 2d) and S₉ = 9/2(2a + 8d) = 9(a + 4d).

   5(a+2d)/9(a+4d) = 5/17 ⇒ 17a + 34d = 9a + 36d ⇒ d = 4a

2. Use the bounds. a₁₅ = a + 14d = a + 56a = 57a. The condition 110 < 57a < 120 allows only one natural number, a = 2, because 57 × 2 = 114. So d = 8.
3. So. S₁₀ = 10/2(2a + 9d) = 5(4 + 72) = 380, which is option (b).

The idea to keep. A ratio of AP sums gives you a relation between a and d, not their values. You need one more fact, here the natural-number bounds, to pin them down.

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