Let S_n denote the sum of the first n terms of an arithmetic progression.

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 $\mathrm{S}_{15}-\mathrm{S}_5$ is equal to :

Accepted Answer

To solve this problem, we will start by using the properties of an arithmetic progression (AP).

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

Given the information:
$S_{10} = 390$

We can plug $n=10$ into the sum formula to get:

$$ S_{10} = \frac{10}{2} (2a + (10-1)d) $$

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

$390 = 10a + 45d$

$$ 78 = 2a + 9d \quad .........\text{(1)} $$

Next, we're given the ratio of the tenth term ($T_{10}$) to the fifth term ($T_5$):
$$ \frac{T_{10}}{T_5} = \frac{15}{7} $$

The $n$th term of an AP is given by:

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

So, for the tenth term:
$T_{10} = a + (10-1)d = a + 9d$

And for the fifth term:

$T_5 = a + (5-1)d = a + 4d$

Now we can write the ratio as:

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

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

$7a + 63d = 15a + 60d$

$63d - 60d = 15a - 7a$

$$ 3d = 8a \quad .........\text{(2)} $$

Now we have two equations (1) and (2):

$$ 78 = 2a + 9d \quad \text{(1)} $$

$$ 3d = 8a \quad \text{(2)} $$

We can solve these equations simultaneously.

From equation (2):

$$ d = \frac{8}{3}a $$

Plugging this back into (1):

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

$78 = 2a + 24a$

$78 = 26a$

$a = 3$

Now we can find $d$:

$$ d = \frac{8}{3}a $$

$$ d = \frac{8}{3} \times 3 $$

$d = 8$

Now we can find $S_{15}$ and $S_5$ using the formula for the sum of an AP.

For $S_{15}$:

$$ S_{15} = \frac{15}{2} (2 \cdot 3 + (15-1) \cdot 8) $$

$$ S_{15} = \frac{15}{2} (6 + 14 \cdot 8) $$

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

$$ S_{15} = \frac{15}{2} \cdot 118 $$

$$ S_{15} = 15 \cdot 59 $$

$S_{15} = 885$

For $S_5$:

$$ S_5 = \frac{5}{2} (2 \cdot 3 + (5-1) \cdot 8) $$

$$ S_5 = \frac{5}{2} (6 + 4 \cdot 8) $$

$$ S_5 = \frac{5}{2} (6 + 32) $$

$$ S_5 = \frac{5}{2} \cdot 38 $$

$$ S_5 = 5 \cdot 19 $$

$S_5 = 95$

The difference $S_{15} - S_{5}$ is:

$S_{15} - S_{5} = 885 - 95$

$S_{15} - S_{5} = 790$

Therefore, the correct answer is Option C, which is 790.

Solutions · 1