The sum of the three no. in A.P(arithmetic progression)is 27

Question

and the sum of their squares is 341?

Answer 1

Let the three numbers in arithmetic sequence = a, (a+d), (a + 2d)

The sum of three terms = a + (a + d )+ (a + 2d)

a + (a + d )+ (a + 2d) = 27

3a + 3d = 27

a + d = 9 ---> (1)

And the sum of their squares a² + (a + d)² + (a + 2d)²

a² + (a + d)² + (a + 2d)² = 341

a² + a² + d² + 2ad +  a² + 4d² + 4ad = 341

3a² + 5d² + 6ad = 341

3a² + 5d² + 6ad - 341 = 0

From (1), a = 9 - d

3(9 - d)² + 5d² + 6(9 - d)d - 341 = 0

3(81 + d² - 18d) + 5d² + 54d - 6d² - 341 = 0

243 + 3d² - 54d + 5d² + 54d - 6d² - 341 = 0

2d²  - 98 = 0

d²  - 49 = 0

d²  = 49

d = ± 7

Consider d = 7 in equation (1).

a + 7 = 9

a = 2

The three numbers are 2, 9, 16.