Question

Suppose a₁, a₂, … a_n, … be an arithmetic progression of natural numbers. If the ration 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

• A. 290
• B. 380
• C. 460
• D. 510

Answer 1

The Correct Option is B

 To solve this problem, we need to use the properties of an arithmetic progression (AP). Let's break down the information given and use it to find the required sum of the first ten terms of the progression.

1. The general formula to find the sum of the first n terms of an arithmetic progression is: \(S_n = \frac{n}{2} \times (2a + (n-1)d)\), where \(a\) is the first term, \(d\) is the common difference, and \(S_n\) is the sum of the first \(n\) terms.
2. According to the problem, the ratio of the sum of the first five terms to the sum of the first nine terms is \(\frac{5}{17}\):

\(\frac{S_5}{S_9} = \frac{5}{17}\)

Substitute \(S_5\) and \(S_9\) using the sum formula: \(\frac{\frac{5}{2} \times (2a + 4d)}{\frac{9}{2} \times (2a + 8d)} = \frac{5}{17}\)

After simplification: \(\frac{5 \times (2a + 4d)}{9 \times (2a + 8d)} = \frac{5}{17}\)

\(17 \times 5 \times (2a + 4d) = 5 \times 9 \times (2a + 8d)\)

Simplifying, we obtain: \(34a + 68d = 18a + 72d\)

This simplifies to: \(16a = 4d \, \Rightarrow \, a = \frac{d}{4}\)

1. The 15th term of the AP, \(a_{15}\), is within the range 110 and 120, giving us: \(110 < a_{15} < 120\).

The 15th term in an AP is \(a_{15} = a + 14d\).

Substitute \(a = \frac{d}{4}\) into the equation: \(a_{15} = \frac{d}{4} + 14d = \frac{d + 56d}{4} = \frac{57d}{4}\)

Thus, \(110 < \frac{57d}{4} < 120\)

Multiplying throughout by 4, we get: \(440 < 57d < 480\)

Solving these inequalities: \(7.72 \approx d < 8.42\) (since \(d\) is a natural number, we choose \(d = 8\))

1. Now, calculate \(a\) using \(a = \frac{d}{4} = 2\).
2. Find the sum of the first 10 terms:

\(S_{10} = \frac{10}{2} \times (2 \times 2 + 9 \times 8) = 5 \times (4 + 72) = 5 \times 76 = 380\)

Therefore, the sum of the first 10 terms of the progression is 380.