Question

find four number in AP such that there sum is 64 and product of extremes is 220​

Answer 1

Answer:-

10 , 13 , 22 , 19

Given:-

Sum = 64

Products of their extremes is 220

To find :-

The four numbers in A. P

Solution:-

Let the four numbers in A. P be a + 2d , a -2d , a + d, a - d

Then,

A/Q.

$(a -2d) +(a +d )+(a-d) +(a+2d) = 64$

$a + a + a +a -2d +2d -d +d = 64$

$4a = 64$

$a = \dfrac{64}{4}$

$a = 16$

Now,

$(a-2d) (a+2d) = 220$

$a^2 - (2d)^2 = 220$

$(16)^2 - 4d^2 = 220$

$-4d^2 = 220-256$

$-4d^2 = -36$

$d^2 = \dfrac{36}{4}$

$d^2 = 9$

$d = \sqrt{9}$

$d = \pm 3$

Now numbers are :-

a - d = 16 - 3 = 13

a + d = 16 + 3 = 19

a -2d = 16 - 2× 3 = 16 - 6 = 10

a +2d = 16 + 2× 3 = 22