Question

Find the three numbers in AP whose sum is 21 and sum of their squares if 179

Answer 1

Answer:

Let the three required numbers be a-d, a ,a+d.

a-d + a + a+d = 21

3a = 21

a = 7 -----------(1)

(a-d)² + a² + (a+d)² = 179

a² - 2ad + d² + a² + a² +2ad + d² = 179           (using (a+b)² and (a-b)²)

3a² + 2d² = 179

3×7² + 2d² = 179                                               (using (1))\

2d² = 179 - 147

2d² = 32

d² = 32/2

d² = 16

∴d = ±8

When d=8:                      

a-d = -1

a+d = 15

a = 7

When d = -8:

a-d = 15

a+d = -1

a = 7

∴ The three numbers are -1, 7 and 15 or  15, 7 and -1.

Hope it helps you!!

Answer 2

Answer:

d) 3,7,11

Explanation:

The sum of the first three number (3, 7, 11) is 21 &

The sum of their Squares ( 9, 49, 121) is 179

So , the three number are 3 ,7, 11