Question

If first, fifth and last terms of an AP is $l,m,p$ respectively and sum of the AP is $\frac{(l+p)(4p-m-5l)}{K(m-l)}$, then $K$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A

$2$

Explanation for the correct option:

Step 1. Find the value of $K$:

Given, $a=l$

$a_5=m$

$\Rightarrow$ $a+4d=m$

$\Rightarrow$ $l+4d=m$

$\Rightarrow$ $4d=m-l$

$\Rightarrow$ $d=\frac{\left(m-l\right)}{4}$

and $a_n=p$

$\Rightarrow$$a+(n-1)d=p$

$\Rightarrow$ $l+(n-1)d=p$

Step 2. Put the value of $d$,

$l+(n-1)\frac{\left(m-l\right)}{4}=p$

$\Rightarrow$ $4(p-l)=(n-1)(m-l)$

$\Rightarrow$ $n-1=\frac{4(p-l)}{(m-l)}$

$\Rightarrow$ $n=\frac{4(p-l)}{(m-l)}+1$

$$\begin{gathered} =\frac{4(p-l)+(m-l)}{(m-l)} \\ =\frac{(4p-m-5l)}{(m-l)} \end{gathered}$$

Step 3. sum of $n$ terms of this AP will be,

$S_n=\left(\frac{n}{2}\right)$(first term + last term)

$=\left[\frac{(4p-m-5l)}{2(m-l)}\right](l+p)$

Compare $S_n$ with $\frac{(l+p)(4p-m-5l)}{K(m-l)}$, we get

$\mathit{K}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}$

Hence, Option ‘A’ is Correct.