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

Answer: (C)

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: $S_{15} - S_5 = 790$.