Question

Is $56$ a term of the A.P.$4,4\frac{1}{2},5,5\frac{1}{2},6,...?$

Answer 1

AP = $4,\frac{9}{2},5,\frac{11}{2},6,...$

Here $a=4$ and

$d=5-\frac{9}{2}=\frac{10-9}{2}=\frac{1}{2}$

Let $56$ be a term, say, $nth$ term of this AP.

We know that

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

So,

$56=4+(n-1)\times \frac{1}{2}$

$\Rightarrow 2\times (56-4)=n-1$

$\Rightarrow 2\times 52=n-1$

$\Rightarrow 104=n-1$

$\Rightarrow 105=n$

Hence, $56$ is the ${105}^{th}$ term of this given AP.