Find four terms in an ap whose sum is 20 and sum of whose squares is 120
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Answer: if a=4, d=2 ∴ the 4 terms are 2, 4, 6 and 8
if a= 6, d= -2 ∴ the 4 terms are 6, 4, 2 and 0
Step-by-step explanation: LET THE 4 TERMS BE (a-d), a, (a+d), (a+2d)
sum of the terms
(a-d)+ a + (a+d) + (a+2d) = 20
a -d + a + a + d + a + 2d =20
4a + 2d = 20
2a + d = 10
d = 10-2a
sum of whole squares
(a-d)² + a² + (a+d)² + (a+2d)² = 120
a² + d² -2ad + a² + a² + d² + 2ad + a² + 4d² + 4ad = 120
4a² + 6d² + 4ad = 120
2a² + 3d² + 2ad = 60
2a² + 3(10-2a)² + 2a(10-2a) = 60
2a² + 3(100+4a²-40a) + 20a - 4a² = 60
2a² + 300 + 12a² - 120a + 20a - 4a² = 60
10a² - 100a + 300 = 60
a² - 10a + 30 = 6
a² - 10 + 24 = 0
a² - 4a - 6a + 24 = 0
(a-4)(a-6) = 0
a=4 a=6
if a=4
d= 10-2a= 10 - 2(4) = 2
∴ the numbers are (a-d), a, (a+d), (a+2d) = (4-2), 4, (4+2), (4+2(2) = 2, 4, 6, 8.
if a=6
d=10-2a= 10 - 2(6) = -2
∴ the numbers are (a-d), a, (a+d), (a+2d)=(4-(-2), 4, (4+(-2), (4+2(-2)= 6, 4, 2, 0.