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

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.