Question

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

Answer 1

Hint: Assume $a,a+d,a+2d,a+3d$, as the four consecutive terms of the A.P. Here ‘a’ is the first term and ‘d’ is the common difference of the A.P. Take the sum of these four numbers and equate this sum equal to 32. Further, take the ratio of product of $a,a+3d$ and $a+d,a+2d$ and equate this with 7:15. Solve the two equations formed and get the values of $a$ and $d$ to get the four numbers.

Complete step-by-step answer:

Let us assume the four consecutive terms of A.P as: $a,a+d,a+2d,a+3d$, here ‘$a$’ is the first term of the A.P and ‘$d$’ is the common difference.
Now, let us come to the question. It is given that the sum of the four terms is 32. Therefore,
$\begin{aligned} & a+a+d+a+2d+a+3d=32 \\\ & 4a+6d=32 \\\ & 2a+3d=16........................(i) \\\ \end{aligned}$
Now, product of first and the last term $=a\times (a+3d)={{a}^{2}}+3ad$
And, product of two middle terms $=(a+d)\times (a+2d)={{a}^{2}}+3ad+2{{d}^{2}}$
It is given that the ratio of these products is 7:15. Therefore,
$\dfrac{{{a}^{2}}+3ad}{{{a}^{2}}+3ad+2{{d}^{2}}}=\dfrac{7}{15}$
By cross-multiplication we get,
$\begin{aligned} & 15{{a}^{2}}+45ad=7{{a}^{2}}+21ad+14{{d}^{2}} \\\ & 8{{a}^{2}}+24ad-14{{d}^{2}}=0 \\\ & 4{{a}^{2}}+12ad-7{{d}^{2}}=0 \\\ \end{aligned}$
Splitting the middle term we get,

& 4{{a}^{2}}+14ad-2ad-7{{d}^{2}}=0 \\\ & 2a(2a+7d)-d(2a+7d)=0 \\\ & (2a+7d)(2a-d)=0 \\\ & \text{either }a=\dfrac{d}{2}\text{ or }a=\dfrac{-7d}{2} \\\ \end{aligned}$$ Case (i): substituting $a=\dfrac{d}{2}$ in equation (i). $\begin{aligned} & \Rightarrow 2\times \dfrac{d}{2}+3d=16 \\\ & \Rightarrow d+3d=16 \\\ & \Rightarrow 4d=16 \\\ & \Rightarrow d=4 \\\ \end{aligned}$ Therefore, $a=\dfrac{d}{2}=\dfrac{4}{2}=2$. Hence, the four terms of the A.P are: 2, 6, 10, 14. Case (ii): Substituting $a=\dfrac{-7d}{2}$ in equation (i). $\begin{aligned} & \Rightarrow 2\times \left( \dfrac{-7d}{2} \right)+3d=16 \\\ & \Rightarrow -7d+3d=16 \\\ & \Rightarrow -4d=16 \\\ & \Rightarrow d=-4 \\\ \end{aligned}$ Therefore, $a=\dfrac{-7d}{2}=\dfrac{(-7)\times (-4)}{2}=14$ Hence, the four terms of the A.P are: 14, 10, 6, 2. Note: It is very important to note that there are two arithmetic progressions that are formed by the above conditions. One of the A.P is increasing and the other is decreasing. We have to consider both the values of $a\text{ and }d$ because both are real and satisfy the conditions of the question.