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.

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.