Divide $56$ in four parts in A.P. such that the ratio of the product of their extremes ( $1^{s t}$ and $4^{t h}$ ) to the product of means ( $2^{n d}$ and $3^{r d}$ ) is $5 : 6$ .

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Solution

Let four terms in $A . P a - 3 d, a - d, a + d, a + 3 d$

Gicen: $\frac{(a - 3 d) (a + 3 d)}{(a - d) (a + d)} = \frac{5}{6}$ ...(1)

$a - 3 d + a - d + a + d + a + 3 d = 56$

$4 a = 56$

$a = 14$

Put $a = 14$ in eq. (1)

$\frac{(14 - 3 d) (14 + 3 d)}{(14 - d) (14 + d)} = \frac{5}{6}$

$\frac{196 - 9 d^{2}}{196 - d^{2}} = \frac{5}{6}$

$6 (196 - 9 d^{2}) = 5 (196 - d^{2})$

$1176 - 54 d^{2} = 980 - 5 d^{2}$

$1176 - 980 = - 5 d^{2} + 54 d^{2}$

$196 = 49 d^{2}$

$\frac{(14)^{2}}{(7)^{2}} = d^{2}$

$d = \pm 2$

A.P when $a = 14$ , $d = 2$

$8, 12, 16, 20$

A.P when $a = 14$ , $d = - 2$

$20, 16, 12, 8$

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Similar questions

56

(1 s t

4 t h)

(2 n d

3 r d)

5 : 6

56

5 : 6

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the prodcut of their means in 5:6.

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