# The sum of the 4th and the 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44.

Question:

The sum of the 4th and the 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44. Find the sum of its first 10 terms.

Solution:

Let a be the first term and d be the common difference of the AP.

$\therefore a_{4}+a_{8}=24$                    (Given)

$\Rightarrow(a+3 d)+(a+7 d)=24 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 2 a+10 d=24$

$\Rightarrow a+5 d=12 \quad \cdots(1)$

Also,

$\therefore a_{6}+a_{10}=44$                                (Given)

$\Rightarrow(a+5 d)+(a+9 d)=44 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 2 a+14 d=44$

$\Rightarrow a+7 d=22 \quad \ldots(2)$

Subtracting (1) from (2), we get

$(a+7 d)-(a+5 d)=22-12$

$\Rightarrow 2 d=10$

$\Rightarrow d=5$

Putting d = 5 in (1), we get

$a+5 \times 5=12$

$\Rightarrow a=12-25=-13$

Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we get

$S_{10}=\frac{10}{2}[2 \times(-13)+(10-1) \times 5]$

$=5 \times(-26+45)$

$=5 \times 19$

$=95$

Hence, the required sum is 95.