Question

The $4^{th}$ term of an A.P. is three times the first and the $7^{th}$ term exceeds twice the third term by $1$. Find the first term and the common difference.

Answer 1

The $4^{th}$ term is three times of the first term.

As we know for AP, the relation will be as followed

$3a=a+3d,$

$2a=3d,$ or

$a=\frac{3}{2}\times d$ …(1)

The $7^{th}$ term exceeds twice the third term by $1$. The relation is

$a+6d=2(a+2d)+1,$

$a+6d=2a+4d+1,$

$2a-a=6d-4d-1,$o

$a=2d-1$ …(2)

Equating equation (1) and (2),

$a=\left(3/2\right)d=2d-1,$

$3d=4d-2,$

$4d-3d=2,$ or

$d=2anda=3.$

The first term is $3$ and the common difference is $2$.

Checking the given conditions,

$3^{th}term=3+2\times 2=7$

$4^{th}term=3+3\times 2=9$

$7^{th}term=3+6\times 2=15$

Hence true.