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 $5 : 6$ .

Open in App

Solution

Let the four parts that are in AP be $a - 3 d, a + d, a + 3 d$

$a - 3 d + a - d + a + d + a + 3 d = 36$

$4 a = 56$

$a = 14$

According to question,

$(a - 3 d) (a + 3 d) \div (a - d) (a + d) = 5 \div 6$

$[a^{2} - 9 d^{2}] \div [a^{2} - d^{2}] = 5 \div 6$

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

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

$49 d^{2} = 196$

$d = + - 2$

Four part. 8,12 if $d = 2$ and. 20,16,12 if $d = - 2$

Suggest Corrections

Similar questions

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.

Join BYJU'S Learning Program