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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 $-13, -8, -3, 2, ......$