Question

The ${17}^{th}$ term of $AP$ is $5$ more than twice it's $8^{th}$ term. If the ${11}^{th}$ term of the AP is $43$, find its $n^{th}$ term

Answer 1

Let ${}^{\prime }{a}^{\prime }$ be the first term & ${}^{\prime }{d}^{\prime }$ be the common difference

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

As per question, we have

$a_{17}=2\times \left(a_8\right)+5$

$\Rightarrow a+(17-1)d=2[a+(8-1\left)d\right]+5$ $[\because a_n=a+(n-1)\times d]$

$\Rightarrow a+16d=2(a+7d)+5$

$\Rightarrow a+16d=2a+14d+5$

$\Rightarrow 2a-a+14d-16d=-5$

$\Rightarrow a-2d=-5\dots \dots \left(i\right)$

Now, It is given that ${11}^{th}$ term is $43.$

$a_{11}=43$

$\Rightarrow a+10d=43\dots \dots \left(ii\right)$

on subtracting $eq.\left(i\right)$ from $eq.\left(ii\right)$, we get

$a+10d-(a-2d)=43-(-5)$

$\Rightarrow 12d=48$

$\Rightarrow d=4$

on putting $d=4$ in $eq.\left(ii\right)$, we get

$a+10\times 4=43$

$\Rightarrow a=43-40=3$

Now, $n^{th}$ term of the given AP is

$n^{th}$ term $=a+(n-1)d$

$=3+(n-1)4$

$=3+4n-4$

$=4n-1$

Hence, its $n^{th}$ term $=4n-1$