Question

The sum of the $4^{th}$ and $8^{th}$ terms of an AP is 24 and the sum of $6^{th}$ and ${10}^{th}$ terms is 44. Find the first three terms of the AP.

Answer 1

The sum of $4^{th}$ and $8^{th}$ terms of an AP is 24 and sum of $6^{th}$ and ${10}^{th}$ terms is 44.

It means $a_4$ + $a_8$ = 24 and $a_6$ + $a_{1}0$ = 44

Using formula $a_n$ = a + (n−1) d, to find $n^{th}$ term of arithmetic progression, we can say that

a+(4−1)d+(a+(8−1)d)=24 and, a+(6−1)d+(a+(10−1)d)=44

$\Rightarrow$ a + 3d + a + 7d=24 and, a+5d+a+9d=44

$\Rightarrow$ 2a + 10d = 24 and, 2a+14d=44

$\Rightarrow$ a+5d=12 and, a+7d=22

These are equations in two variables. Let’s solve them using method of substitution.

Using equation a+5d=12, we can say that a=12−5d (1)

Putting (1) in equation a+7d=22, we can say that

12−5d+7d=22

$\Rightarrow$12+2d=22

$\Rightarrow$2d=10

$\Rightarrow$d = $\frac{10}{2}$ = 5

Putting value of d in equation: a=12−5d, we get

a=12−5 (5) =12−25=−13

Therefore, first term =a=−13

And, Common difference =d=5

Therefore, AP is -13, -8, -3, 2...

Its first three terms are -13, -8 and -3.