Question

Four numbers are in arithmetic progression. The sum of the first and last term is $8$ and the product of both middle terms is $15$. The least number of the series is:

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for correct option:

Step: 1 Find out the all four numbers of the series.

Let, the series of four numbers be $a-3d$, $a-d$, $a+d$, and $a+3d$.

Given, the sum of the first term and the last term is $8$.

$\Rightarrow$ $a-3d+a+3d=8$

$\Rightarrow$ $2a=8$

$\Rightarrow$ $a=4$

Again, the product of two middle terms is $15$.

$\Rightarrow$$(a-d)(a+d)=15$

$\Rightarrow$ $a^2-d^2=15$

$\Rightarrow$ $4^2-d^2=15$

$\Rightarrow$ $d^2=1$

$\Rightarrow$ $d=1$

Step:2 Find out the least number of the series.

$\therefore$four numbers of the series are:

$a-3d$, $a-d$, $a+d$, and $a+3d$

$=1,3,5,7$

Thus, the least number of the series is $\mathbf{1}$.

Hence, Option (D) is the correct answer.