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

Given:

\[ S_{10} = 390, \quad \text{Ratio of the 10th and 5th terms} = \frac{T_{10}}{T_5} = \frac{15}{7} \]

Let \( a \) be the first term and \( d \) be the common difference of the arithmetic progression.

The sum of the first \( n \) terms of an AP is given by:

\[ S_n = \frac{n}{2}[2a + (n - 1)d] \]

For \( n = 10 \):

\[ S_{10} = \frac{10}{2}[2a + 9d] = 5(2a + 9d) = 390 \] \[ 2a + 9d = 78 \quad \text{(1)} \]

The 10th term \( T_{10} \) and 5th term \( T_5 \) are given by:

\[ T_{10} = a + 9d, \quad T_5 = a + 4d \]

Given:

\[ \frac{T_{10}}{T_5} = \frac{a + 9d}{a + 4d} = \frac{15}{7} \]

Cross-multiplying:

\[ 7(a + 9d) = 15(a + 4d) \]

Expanding:

\[ 7a + 63d = 15a + 60d \]

Rearranging terms:

\[ 8a = 3d \implies d = \frac{8a}{3} \quad \text{(2)} \]

Substituting \( d = \frac{8a}{3} \) into equation (1):

\[ 2a + 9\left(\frac{8a}{3}\right) = 78 \]

Multiplying through by 3:

\[ 6a + 72a = 234 \]

Combining terms:

\[ 78a = 234 \implies a = 3 \]

Substituting \( a = 3 \) back into equation (2):

\[ d = \frac{8 \times 3}{3} = 8 \]

Finding \( S_{15} \) and \( S_5 \):

\[ S_{15} = \frac{15}{2}[2a + 14d] = \frac{15}{2}[2 \times 3 + 14 \times 8] = \frac{15}{2}[6 + 112] = \frac{15}{2} \times 118 = 885 \] \[ S_5 = \frac{5}{2}[2a + 4d] = \frac{5}{2}[2 \times 3 + 4 \times 8] = \frac{5}{2}[6 + 32] = \frac{5}{2} \times 38 = 95 \]

Calculating \( S_{15} - S_5 \):

\[ S_{15} - S_5 = 885 - 95 = 790 \]

Conclusion: \( S_{15} - S_5 = 790 \).