Question

Divide 56 into 4 parts which are in AP such that the ratio of product of extremes to the product of means is 5 : 6.


Answer with complete steps and reasoning.


Class 10 | CBSE | Arithmetic Progression | 3 Marks Question

Answer 1

HELLO DEAR,

Let the four parts be (a – 3d), (a – d), (a + d) and (a + 3d).

given that :-

SUM = 56

(a – 3d) + (a – d) + (a + d) + (a + 3d) = 56

=> a - 3d + a - d + a +d +a + 3d = 56

=> 4a = 56

=> a =56/4

=> a = 14

AND,

Also ratio is GIVEN THAT:- 5:6

$\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{5}{6} \\ \\ = > \frac{ {a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} } = \frac{5}{6} \\ \\ = > \frac{ {14}^{2} - 9 {d}^{2} }{ {14}^{2} - {d}^{2} } = \frac{5}{6} \: \: \: \: \: \: ( {a = 14}) \\ \\ = > \frac{196 - 9 {d}^{2} }{196 - {d}^{2} } = \frac{5}{6} \\ \\ = > 1176 - 54 {d}^{2} = 980 - 5 {d}^{2} \\ \\ = > 1176 - 980 = 54 {d}^{2} - 5 {d}^{2} \\ \\ = > 49 {d}^{2} = 196 \\ \\ = > {d}^{2} = 4 \\ \\ = > d = + - 2$


if d = +2

THEN,

(a – 3d), (a – d), (a + d) and (a + 3d).

=> (8 ,12,16,20)

if d = -2

THEN,

(a – 3d), (a – d), (a + d) and (a + 3d).

=> (20 ,16 ,12 8)


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

hy
your answer in this pic