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Divide 32 into four parts which are in A.P. such that the prod

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Divide $32$ into four parts which are in A.P. such that the product of extremes is to the product of means is $7 : 15$ .

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Divide

32

into four parts which are in A.P. such that the product of extremes is to the product of means is

7 : 15

11 mins ago

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11 mins ago

Text solutionVerified

Let the four parts be

(a - 3 d), (a - d), (a + d)

and

(a + 3 d)

Then, Sum of the numbers

= 32

⟹ (a - 3 d) + (a - d) + (a + d) + (a + 3 d) = 32 ⟹ 4 a = 32 ⟹ a = 8

It is given that

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

⟹ \frac{a ^{2} - 9 d ^{2}}{a ^{2} - d ^{2}} = \frac{7}{15}

⟹ \frac{64 - 9 d ^{2}}{64 - d ^{2}} = \frac{7}{15} ⟹ 128 d^{2} = 512 ⟹ d^{2} = 4 ⟹ d = \pm 2

Thus, the four parts are

a - 3 d, a - d, a + d

and

3 d

, i.e.

2, 6, 10, 14

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Question Text	Divide $32$ into four parts which are in A.P. such that the product of extremes is to the product of means is $7 : 15$ .
Answer Type	Text solution:1
Upvotes	150