Question

Divide 96 into four parts which are in A.P. an the ratio between product of their means to product of their extremes is 15: 7.

Answer 1

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Answer 2

Answer:

The four parts which are in AP when 96 is divided is 6, 18, 30, 42.

Step-by-step explanation:

Let a-3d, a-d, a+d and a+3d be the four equal parts in AP

Then we get a-3d+a-d+a+d+a+3d=96

4a=96

a=96/4

a=24

Given product of means /product of extremes=15/7

From the given question we have $\frac{[(a-3 d)+(a+3 d)]}{[(a-d)+(a+d)]}=\frac{7}{15}$

Simplifying the above equation we get

$\frac{a^{2}-d^{2}}{(a-3 d)(a+3 d)}=\frac{15}{7}$

By applying the $a^{2}-b^{2}$ to the denominator of the above expression

$\frac{a^{2}-d^{2}}{a^{2}-(3 d)^{2}}=\frac{15}{7}$

Cross multiply the above expression

We get

$\begin{array}{l}{7\left(a^{2}-d^{2}\right)=15\left(a^{2}-9 d^{2}\right)} \\ {7 a^{2}-7 d^{2}=15 a^{2}-135 d^{2}} \\ {7 a^{2}-15 a^{2}=-135 d^{2}+7 d^{2}} \\ {-8 a^{2}=-128 d^{2}}\end{array}$

Multiply throughout by (-)

$\begin{array}{l}{8 a^{2}=128 d^{2}} \\ {d^{2}=8 a^{2} / 128}\end{array}$

We know the value of a so substitute in the above expression

$d^{2}=8(24)^{2} / 128$ =468/128

$d^{2}=36$

d=√36

$\mathrm{d}=\pm 6$

Therefore

Case 1: a=24 d=6

a-3d=24-18=6

a-d=24-6=18

a+3d=24+18=42

a+d=24+6=30

Case 2: a=24 d=-6

a-3d=24+18=42

a-d=24+6=30

a+3d=24-18=6

a+d=24-6=18

Therefore the required terms are 6, 18, 30, 42