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 solve the given problem, we need to find the value of \( S_{15} - S_{5} \) for an arithmetic progression (AP) where \( S_{10} = 390 \) and the ratio of the tenth term to the fifth term is \( 15:7 \).

The sum of the first \( n \) terms of an arithmetic progression is given by the formula:

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

where \( a \) is the first term and \( d \) is the common difference of the AP.

Given \( S_{10} = 390 \), we can set up the equation:

\(\frac{10}{2} (2a + 9d) = 390\)

This simplifies to:

\(5(2a + 9d) = 390\)

\(2a + 9d = 78 \quad \ldots (1)\)

The nth term of an AP is given by:

\(T_n = a + (n - 1)d\)

Given the ratio of the tenth term to the fifth term is \( 15:7 \), we have:

\(\frac{T_{10}}{T_{5}} = \frac{15}{7}\)

This implies:

\(\frac{a + 9d}{a + 4d} = \frac{15}{7}\)

Cross-multiplying, we get:

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

This simplifies to:

\(7a + 63d = 15a + 60d\)

\(8a = 3d \quad \ldots (2)\)

Now, solve equations (1) and (2) simultaneously:

From equation (2):

\(a = \frac{3}{8}d\)

Substituting this into equation (1):

\(2\left(\frac{3}{8}d\right) + 9d = 78\)

\(\frac{3}{4}d + 9d = 78\)

\(\frac{3d}{4} + 9d = 78\)

\(\frac{3d + 36d}{4} = 78\)

\(39d = 312\)

\(d = 8\)

Substitute \( d = 8 \) back into the equation \( a = \frac{3}{8}d \):

\(a = \frac{3}{8} \times 8 = 3\)

Now calculate \( S_{15} \) and \( S_{5} \):

\(S_{15} = \frac{15}{2} (2a + 14d)\)

\(S_{15} = \frac{15}{2} (2 \times 3 + 14 \times 8)\)

\(S_{15} = \frac{15}{2} (6 + 112)\)

\(S_{15} = \frac{15}{2} \times 118 = 885\)

\(S_{5} = \frac{5}{2} (2a + 4d)\)

\(S_{5} = \frac{5}{2} (2 \times 3 + 4 \times 8)\)

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

Therefore,

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

Therefore, the answer is 790.

Answer 2

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