Question

The sum of four consecutive numbers in an A.P. is $32$ and the ratio of the product of the first and the last term to the product of two middle terms is $7:15$. Find the numbers.

Answer 1

The objective is to find the four consecutive terms of an A.P. The sum of those terms is $32$ and the ratio of the products of the first and last term to the middle terms is $7:15$.

Step 1 :

Suppose the four terms of the A.P. are $\left(a-3d\right),\left(a-d\right),\left(a+d\right),$ and $\left(a+3d\right)$.

As their sum is $32$ then,

$$\begin{gathered} \Rightarrow a-3d+a-d+a+d+a+3d=32 \\ \Rightarrow 4a=32 \\ \Rightarrow a=8 \end{gathered}$$

Step 2 :

Using the second condition,

$$\begin{gathered} \Rightarrow \frac{\left(a-3d\right)\left(a+3d\right)}{\left(a-d\right)\left(a+d\right)}=\frac{7}{15} \\ \Rightarrow \frac{a^2+3ad-3ad-9d^2}{a^2+ad-ad-d^2}=\frac{7}{15} \\ \Rightarrow \frac{a^2-9d^2}{a^2-d^2}=\frac{7}{15} \end{gathered}$$

Applying the value of $a$,

$$\begin{gathered} \Rightarrow 15\left(64-9d^2\right)=7\left(64-d^2\right) \\ \Rightarrow 960-135d^2=448-7d^2 \\ \Rightarrow 128d^2=512 \\ \Rightarrow d^2=4 \end{gathered}$$

$d=\pm 2$

Step 3:

The possible four consecutive terms of an A.P.

$\begin{array}{c}\text{Take,}d=2\\ \therefore (a-3d)=(8-3\times 2)\\ =8-6\\ =2\\ \therefore (a-d)=8-2\\ =6\\ \therefore (a+d)=8+2\\ =10\\ \therefore (a+3d)=8+3\times 2\\ =8+6\\ =14\end{array}$

Now $\text{take,}d=-2$

$\begin{array}{c}\therefore (a-3d)=(8-3\times (-2))\\ =8+6\\ =14\\ \therefore (a-d)=8-(-2)\\ =8+2\\ =10\\ \therefore (a+d)=8+(-2)\\ =8-2\\ =6\\ \therefore (a+3d)=8+3(-2)\\ =8-6\\ =2\end{array}$

Final Answer :

The terms can be $2,6,10,$ and $14$ or $14,10,6,$ and $2$.