Question

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

Answer 1

The nth term of an A.P is given by

$a_n=a_1+(n-1)d$

4th term of A.P is given by

$a_4=a_1+(4-1)d$

$a_4=a_1+3d...eq\left(1\right)$

8th term of A.P is given by

$a_8=a_1+(8-1)d$

$a_8=a_1+7d...eq\left(2\right)$

given that sum of 4th and 8th term is 24 $⟹a_4+a_8=24$

from eq(1) and eq(2)

$a_1+3d+a_1+7d=24$

$⟹2a_1+10d=24\dots ...eq\left(3\right)$

6th term of A.P is given by

$a_6=a_1+(6-1)d$

$a_6=a_1+5d...eq\left(4\right)$

10th term of A.P is given by

$a_{10}=a_1+(10-1)d$

$a_{10}=a_1+9d...eq\left(5\right)$

given that sum of 6th and 10th term is 34 $⟹a_6+a_{10}=34$

from eq(4) and eq(5)

$a_1+5d+a_1+9d=34$

$⟹2a_1+14d=34\dots ...eq\left(6\right)$

substract eq(3) by eq(6) we get

$⟹2a_1-2a_1+14d-10d=34-24$

$⟹4d=10$

$⟹d=\frac{10}{4}$

$⟹d=\frac{5}{2}...\dots ..eq\left(7\right)$ common difference.

put value of d=2.5 in eq(3) we get

$⟹2a_1+10\times 2.5=24$

$⟹2a_1+25=24$

$⟹2a_1=24-25$

$⟹2a_1=-1$

$⟹a_1=-\frac{1}{2}=$ first term of A.P