Question

Find three numbers in
a.p. such that their sum is 27 and the sum of their squares is 341

Answer 1

let the number. be...
x,x-d,x+d....
so ..their sum =3x=27
so x=9..
x^2+(x-d)^2+(x+d)^2=3x^2+2d^2
by putting value of x..we got
243+2d^2=341
so 2d^2=98
d^2=49
d=7
so numbers are 2,9,16

Answer 2

The three numbers are 2, 9 and 16.

Step-by-step explanation:

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

According to the question,

a - d + a + a + d = 27 ⇒ 3a = 27 ⇒ a = 9

and (a - d)² + a² + (a + d)² = 341

⇒ (9 - d)² + 9² + (9 + d)² = 341

⇒ 81 - 18d + d² + 81 + 81 + 18d + d² = 341

⇒ 2d² + 243 = 341

⇒ 2d² = 98

⇒ d² = 49

⇒ d = ± 7

• When a = 9 and d = + 7, the three numbers are

9 - (+ 7), 9 and 9 - (- 7) ⇒ 2, 9 and 16

• When a = 9 and d = - 7, the three numbers are

9 - (- 7), 9 and 9 + (- 7) ⇒ 16, 9 and 2

∴ the required three numbers are 2, 9 and 16.

#SPJ3