The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34

Question:

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.

Solution:

In the given problem, the sum of 4th and 8th term is 24 and the sum of 6th and 10th term is 34.

We can write this as,

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

$a_{6}+a_{10}=34$......(2)

We need to find a and d

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 4th 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 $10^{\text {th }}$ term $(n=10)$,

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

$=a+9 d$

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

$(a+5 d)+(a+9 d)=34$

$2 a+14 d=34$.......(4)

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

$(2 a+14 d)-(2 a+10 d)=34-24$

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

$4 d=10$

$d=\frac{10}{4}$

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

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

$a+10\left(\frac{5}{2}\right)=24$

$2 a+5(5)=24$

 

$2 a+25=24$

$2 a=24-25$

On further simplifying, we get,

$2 a=-1$

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

Therefore, for the given A.P $a=\frac{-1}{2}$ and $d=\frac{5}{2}$

 

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