Find four numbers in AP whose sum is 21 and whose sum of squares is 120
Answers
Answer: Let the four numbers in A.P be a-3d, a-d,a+d,a+3d. ---- (1)
Given that Sum of the terms = 20.
= (a-3d) + (a-d) + (a+d) + (a+3d) = 20
4a = 20
a = 5. ---- (2)
Given that sum of squares of the term = 120.
= (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 120
= (a^2 + 9d^2 - 6ad) + (a^2+d^2-2ab) + (a^2+d^2+2ad) + (a^2+9d^2+6ad) = 120
= 4a^2 + 20d^2 = 120
Substitute a = 5 from (2) .
4(5)^2 + 20d^2 = 120
100 + 20d^2 = 120
20d^2 = 20
d = +1 (or) - 1.
Since AP cannot be negative.
Substitute a = 5 and d = 1 in (1), we get
a - 3d, a-d, a+d, a+3d = 2,4,6,8.
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Step-by-step explanation:
I have solved for a and d in provided solution you can find out desire result with the help of first term and difference between terms .
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