Question

Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is $15 : 7$, then $S_{15} - S_{5}$ is equal to:

• A. 800
• B. 890
• C. 790
• D.
  690

Answer 1

The Correct Option is C

To determine the value of \( S_{15} - S_{5} \) for a given arithmetic progression (AP), we are provided with \( S_{10} = 390 \) and the ratio of the tenth term to the fifth term as \( 15:7 \).

The sum of the first \( n \) terms of an AP is calculated using the formula: \(S_n = \frac{n}{2} (2a + (n - 1)d)\), where \( a \) is the first term and \( d \) is the common difference.

Using the given \( S_{10} = 390 \), we establish the equation: \( \frac{10}{2} (2a + 9d) = 390 \), which simplifies to \( 5(2a + 9d) = 390 \), further reducing to \( 2a + 9d = 78 \quad \ldots (1) \).

The \( n \)th term of an AP is defined by \( T_n = a + (n - 1)d \).

The condition that the ratio of the tenth term to the fifth term is \( 15:7 \) translates to: \( \frac{T_{10}}{T_{5}} = \frac{15}{7} \). This implies \( \frac{a + 9d}{a + 4d} = \frac{15}{7} \).

Cross-multiplication yields \( 7(a + 9d) = 15(a + 4d) \), which simplifies to \( 7a + 63d = 15a + 60d \), and finally \( 8a = 3d \quad \ldots (2) \).

Solving equations (1) and (2) simultaneously, we derive \( a = \frac{3}{8}d \) from equation (2). Substituting this into equation (1) gives \( 2\left(\frac{3}{8}d\right) + 9d = 78 \).

This simplifies to \( \frac{3}{4}d + 9d = 78 \), or \( \frac{3d + 36d}{4} = 78 \), resulting in \( 39d = 312 \), so \( d = 8 \).

Substituting \( d = 8 \) back into \( a = \frac{3}{8}d \) gives \( a = \frac{3}{8} \times 8 = 3 \).

Next, we calculate \( S_{15} \) and \( S_{5} \):

\( S_{15} = \frac{15}{2} (2 \times 3 + 14 \times 8) = \frac{15}{2} (6 + 112) = \frac{15}{2} \times 118 = 885 \).

\( S_{5} = \frac{5}{2} (2 \times 3 + 4 \times 8) = \frac{5}{2} (6 + 32) = \frac{5}{2} \times 38 = 95 \).

Therefore, \( S_{15} - S_{5} = 885 - 95 = 790 \).

The final answer is 790.