The sum of the 4th term and the 8th term of an AP is 24. Similarly, the sum of its 6th term and 10th term is 34. Find $ a $ and $ d $ for this AP.

Given:

The sum of the 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 \( a \) and \( d \) for this AP.

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($a$) and the common difference($d$) of the given A.P. are $\frac{-1}{2}$ and $\frac{5}{2}$ respectively.