# Divide 32 into four parts which are the four terms of an AP such that the product of

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.

Solution:

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

$(a-3 d)+(a-d)+(a+d)+(a+3 d)=32$

$\Rightarrow 4 a=32$

$\Rightarrow a=8 \quad \ldots \ldots(1)$

Also,

$(a-3 d)(a+3 d):(a-d)(a+d)=7: 15$

$\Rightarrow \frac{(8-3 d)(8+3 d)}{(8-d)(8+d)}=\frac{7}{15}$         [From (1)]

$\Rightarrow \frac{64-9 d^{2}}{64-d^{2}}=\frac{7}{15}$

$\Rightarrow 15\left(64-9 d^{2}\right)=7\left(64-d^{2}\right)$

$\Rightarrow 960-135 d^{2}=448-7 d^{2}$

$\Rightarrow 135 d^{2}-7 d^{2}=960-448$

$\Rightarrow 128 d^{2}=512$

$\Rightarrow d^{2}=4$

$\Rightarrow d=\pm 2$

When a = 8 and d = 2,

$a-3 d=8-3 \times 2=8-6=2$

$a-d=8-2=6$

$a+d=8+2=10$

$a+3 d=8+3 \times 2=8+6=14$

When a = 8 and d = −2,

$a-3 d=8-3 \times(-2)=8+6=14$

$a-d=8-(-2)=8+2=10$

$a+d=8-2=6$

$a+3 d=8+3 \times(-2)=8-6=2$

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