Divide 15 into three parts which are in A.P. and the sum of their squares is 83.

Let the middle term = a and the common difference = d

The first term = a – d and the succeeding term = a + d

So, the three parts are a – d, a, a + d

According to the question,

Sum of these three parts = 15

⇒ a – d + a + a + d = 15

⇒ 3a = 15

⇒ a = 5

and the sum of their squares = 83

⇒ (a – d)² + a² + (a + d)² = 83

⇒ (5 – d)² + (5)² + ( 5 + d)² = 83 [from(i)]

⇒ 25 + d² – 10d + 25 + 25 + d² + 10d = 83

⇒ 75 + 2d² = 83

⇒ 2d² = 83 – 75

⇒ 2d² = 8

⇒ d² = 4

⇒ d = √4

⇒ d = ±2

Case: (i) If d = 2, then

a – d = 5 – 2 = 3

a = 5

a + d = 5 + 2 = 7

Hence, the three parts are

3, 5, 7

Case: (ii) If d = – 2, then

a – d = 5 – ( – 2) = 7

a = 5

a + d = 5 + ( – 2) = 3

Hence, the three parts are

7, 5, 3