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

Given:

The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44.

To do:

We have to find the A.P.

Solution:

Let the first term of the A.P. be $a$ and the common difference be $d$.

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

$a_{4}=a+(4-1)d$

$=a+3d$......(i)

$a_{8}=a+(8-1)d$

$=a+7d$......(ii)

According to the question,

$a_4+a_8=a+3d+a+7d$

$24=2a+10d$

$24=2(a+5d)$ 

$12=a+5d$

$a=12-5d$......(iii)

$a_{6}=a+(6-1)d$

$=a+5d$......(iv)

$a_{10}=a+(10-1)d$

$=a+9d$......(v)

According to the question,

$a_6+a_{10}=a+5d+a+9d$

$44=2a+14d$

$44=2(a+7d)$ 

$22=a+7d$

$7d=22-(12-5d)$        (From (iii))

$7d=22-12+5d$

$7d-5d=10$

$2d=10$

$d=\frac{10}{2}$

$d=5$

This implies,

$a=12-5(5)$

$a=12-25$

$a=-13$

Therefore,

$a_1=-13$

$a_2=a+d=-13+5=-8$

$a_3=a+2d=-13+2(5)=-13+10=-3$

$a_4=a+3d=-13+3(5)=-13+15=2$

Hence, the required A.P. is $-13, -8, -3, 2, ......$