Question

$4^{th}$ term of an AP is equal to 3 times the first term and $7^{th}$ term exceeds twice the 3rd term by $1.$ Find the first term and the common difference.

Answer 1

We know that the $n^{th}$ of an AP is given by $a_n=a+(n-1)d$

First term $=a$

$4^{th}$ term $=3a$ [Given]

$\Rightarrow a+3d=3a$

$\Rightarrow 3d=2a\dots \dots \left(i\right)$

And, $2\times (3^{rd}$ term) $=7^{th}$ term $-1$ [Given]

$\Rightarrow 2(a+2d)=(a+6d)-1$

$\Rightarrow 2a+4d=a+6d-1$

$\Rightarrow 2a-a+4d-6d=-1$

$\Rightarrow a-2d=-1$

$\Rightarrow \frac{3d}{2}-2d=-1$ [Using eq. (i)]

$\Rightarrow 3d-4d=-2$

$\Rightarrow -d=-2$

$\Rightarrow d=2$

Now, $a=\frac{3d}{2}$

$\therefore a=3$

Hence, the AP is $3,5,7,9,\mathrm{11............}$