Question

Divide 32 into four parts which are the four terms of an AP such that the product of the first and the fourth term is to the product of the second and the third term is 7 : 15. [CBSE 2014]

Answer 1

Let the four parts in AP be (a − 3d), (a − d), (a + d) and (a + 3d). Then,

$$\begin{gathered} \left(a-3d\right)+\left(a-d\right)+\left(a+d\right)+\left(a+3d\right)=32 \\ \Rightarrow 4a=32 \\ \Rightarrow a=8.....\left(1\right) \end{gathered}$$

Also,
$$\begin{gathered} \left(a-3d\right)\left(a+3d\right):\left(a-d\right)\left(a+d\right)=7:15 \\ \Rightarrow \frac{\left(8-3d\right)\left(8+3d\right)}{\left(8-d\right)\left(8+d\right)}=\frac{7}{15}\left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow \frac{64-9d^2}{64-d^2}=\frac{7}{15} \\ \Rightarrow 15\left(64-9d^2\right)=7\left(64-d^2\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow 960-135d^2=448-7d^2 \\ \Rightarrow 135d^2-7d^2=960-448 \\ \Rightarrow 128d^2=512 \\ \Rightarrow d^2=4 \end{gathered}$$
$\Rightarrow d=\pm 2$

When a = 8 and d = 2,
$$\begin{gathered} a-3d=8-3\times 2=8-6=2 \\ a-d=8-2=6 \\ a+d=8+2=10 \\ a+3d=8+3\times 2=8+6=14 \end{gathered}$$

When a = 8 and d = −2,
$$\begin{gathered} a-3d=8-3\times \left(-2\right)=8+6=14 \\ a-d=8-\left(-2\right)=8+2=10 \\ a+d=8-2=6 \\ a+3d=8+3\times \left(-2\right)=8-6=2 \end{gathered}$$

Hence, the four parts are 2, 6, 10 and 14.