Question

Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

Answer 1

Let the numbers be (a – 3d), (a ­– d), (a + d), (a + 3d),

Then,

Sum = 20

(a – 3d) + (a – d) + (a + d) + (a + 3d) = 20

4a = 20

a = 5

Sum of the squares = 120

${(a-3d)}^2$ + ${(a-d)}^2$ + ${(a+d)}^2$ + ${(a+3d)}^2$ = 120

4$a^2$ + 20$d^2$ = 120

$a^2$ + 5 $d^2$ = 30

25 + 5$d^2$ = 30 [Since, a = 5]

5$d^2$ = 5

d = ± 1

If d = 1, then the numbers are 2, 4, 6, 8. If d = – 1, then the numbers are 8, 6, 4, 2.

Thus, the numbers are 2, 4, 6, 8 or 8, 6, 4, 2.