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

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 $-13, -8, -3, 2, ......$  

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Updated on: 10-Oct-2022

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