Q. 135.0( 3 Votes )

Four numbers are in arithmetic progression. If their sum is 20 and the sum of their squares is 120, then find the numbers.

Answer :

Let the first term be a – 3d and common difference be 2d.
Thus arithmetic progression will be a - 3d, a - d, a + d, a+3d.


Given, Sum of terms = 20
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 20
4a = 20
a = 5 …(i)


Also,


Sum of their square = 120


(a – 3d)2 + (a – d)2 + (a + d)2 + (a + 3d)2 = 120


(5 – 3d)2 + (5 – d)2 + (5 + d)2 + (5 + 3d)2 = 120


(25 + 9d2 – 30d) + (25 + d2 – 10d) + (25 + d2 + 10d) + (25 + 9d2 + 30d) = 120


100 + 20d2 = 120


⇒ 20d2 = 20


⇒ d2 = 1


⇒ d = ± 1


If d = 1


First term, a – 3d = 5 – 3 = 2 and common difference, 2d = 2 and four terms will be 2, 4, 6, 8


If d = -1


First term, a – 3d = 5 –(– 3) = 8 and common difference, 2d = -2 and four terms will be 8, 6, 4 and 2.


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