The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of 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 34.

To do:

We have to find the first term and the common difference of 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,

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

$34=2a+14d$

$34=2(a+7d)$ 

$17=a+7d$

$7d=17-(12-5d)$        (From (iii))

$7d=17-12+5d$

$7d-5d=5$

$2d=5$

$d=\frac{5}{2}$

This implies,

$a=12-5(\frac{5}{2})$

$a=12-\frac{25}{2}$

$a=\frac{12\times2-25}{2}$

$a=\frac{24-25}{2}$

$a=\frac{-1}{2}$

Hence, the first term and the common difference of the given A.P. are $\frac{-1}{2}$ and $\frac{5}{2}$ respectively.