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

Solution:

In the given problem, the sum of 4th and 8th term is 24 and the sum of 6th and 10th term is 44. We have to find the A.P

We can write this as,

$a_{4}+a_{8}=24$............(1)

$a_{6}+a_{10}=44$..........(2)

We need to find the A.P

For the given A.P., let us take the first term as a and the common difference as d

As we know,

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

For $4^{\text {th }}$ term $(n=4)$,

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

$=a+3 d$

For 8th term (n = 8),

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

$=a+7 d$

So, on substituting the above values in (1), we get,

$(a+3 d)+(a+7 d)=24$

$2 a+10 d=24$........(3)

Also, for 6th term (n = 6),

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

$=a+5 d$

For 10th term (n = 10),

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

$=a+9 d$

So, on substituting the above values in (2), we get,

$(a+5 d)+(a+9 d)=44$

$2 a+14 d=44$.........(4)

Next we simplify (3) and (4). On subtracting (3) from (4), we get,

$(2 a+14 d)-(2 a+10 d)=44-24$

$2 a+14 d-2 a-10 d=20$

$4 d=20$

$d=\frac{20}{4}$

$d=5$

Further, using the value of d in equation (3), we get,

$a+10(5)=24$

 

$2 a+50=24$

$2 a=24-50$

$2 a=-26$

$a=-13$

So here, $a=-13$ and $d=5$

Therefore, the A.P. is $-13,-8,-3,2, \ldots .$