Find three numbers in AP whose sum is 18 and sum of their square is 116.
Answers
Answer:
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Step-by-step explanation:
Four numbers in AP.whose sum is 20 and sum of their squares is 180.find the numbers.?
Let the numbers be a-2d, a-d, a and a+d. Their sum is
a-2d+ a-d+ a + a+d = 4a-2d = 20, or
2a - d = 10 …(1)
(a-2d)^2+ (a-d)^2+ a^2 + (a+d)^2 = 180, or
a^2–4ad+4d^2+a^2–2ad+d^2+a^2+a^2+2ad+d^2 = 180, or
a^2–4ad+4d^2+a^2+d^2+a^2+a^2+d^2 = 180, or
4a^2–4ad+6d^2 = 180, or
2a^2–2ad+3d^2 = 90 …(2)
From (1) d = 2a -10. Put that in (2) to get
2a^2 -2a(2a–10) + 3(2a–10)^2 = 90, or
2a^2+20a-4a^2+3(4a^2-40a+100) = 90, or
2a^2+20a-4a^2+300–120a+12a^2 = 90, or
10a^2–100a+210 = 0, or
a^2–10a+21=0
(a-3)(a-7) = 0
Hence a = 3 or 7 and the corresponding value of d from (1)
d=2a - 10 is -4 or 4.
So the 4 terms of the AP are 3-[2*(-4)] = 11, 7, 3 and -1, or
-1, 3, 7 and 11.
Check: 11+7+3–1 = 20. Correct.
1^2+3^2+7^2+11^2 = 1+9+49+121 = 180