Question

The sum of four numbers in A.P is 4 and their product is 385. Find the number.

Answer 1

let no. be a-3d,a-d,a+d,a+3d
a+3d+a-3d+a-d+a+d=4
4a=4
a=1
(a+3d)*(a-3d)(a+d)*(a-d)=385
(1+3d)(1-3d)(1+d)(1-d)=385
(1-9d^2)(1-d^2)=385
1-9d^2-d^2+9d^4=385
9d^4-10d^2+1=385
9d^4-10d^2-384=0
factorise it

Answer 2

Answer:

Let the four terms in AP be a-2d , a-d , a , a+d

Now we are given that The sum of four numbers in A.P is 4

So, $a-2d+a-d+a+a+d= 4$

$4a-2d= 4$

$2a-d= 2$ --- 1

We are also given that their product is 385.

$(a-2d)(a-d)(a)(a+d)= 4$

$(a-2d)(a^3-ad^2)= 4$

$(a^4-a^2d^2-2a^3d+2ad^3)= 4$

Substitute the value of d from 1

$(a^4-a^2(2a+2)^2-2a^3(2a+2)+2a(2a+2)^3)= 4$

$(a^4-a^2(2a+2)^2-2a^3(2a+2)+2a(2a+2)^3)= 4$

Solving $a=\frac{-5}{3} , \frac{11}{3}$

At a = -5/3

$2(\frac{-5}{3})-d= 2$

$d=\frac{-16}{3}$

At a = 11/3

$2(\frac{11}{3})-d= 2$

$d=\frac{16}{3}$

At a = -5/3 and d = -16/3

a-2d= 9

a-d=11/3

a= -5/3

a+d=-7

At a = 11/3 and d = 16/3

a-2d= -7

a-d=-5/3

a= 11/3

a+d=9