Question

Find the ${51}^{th}$ term of an AP whose term is $1^{st}$ term is 112 and $3^{rd}$ is 124?

Answer 1

We know that

$T_n=a+(n-1)d$

$T_1=112$

$112=a+(1-1)\times d$

$112=a+0\times d$

$112=a$

$T_3=124$

$124=a+(3-1)\times d$

$124=a+2d\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\left(i\right)$

Putiing the value of ${}^{\prime \prime }{a}^{\prime \prime }$ in equation $1st$

$124=112+2d$

$124-112=2d\Rightarrow 2d=12$

$d=6$

${T_{51}}^{th}$ term

$T_{51}=a+(51-1)d$

$=112+\left(50\right)\times 6$

$=112+300$

$T_{51}=412$