Four numbers are in A.P. If their sum is 20 and the sum of their squares is 120, then the middle terms are

2,4

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4,6

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6,8

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8,10

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Solution

The correct option is B
4,6

Let the numbers are $a - 3 d, a - d, a + d, a + 3 d$
Given: $a - 3 d + a — d + a + d + a + 3 d = 20$
$4 a = 20$
$a = 5$ and
${(a - 3 d)}^{2} + {(a - d)}^{2} + {(a + 3 d)}^{2} + {(a + 3 d)}^{2} = 120$
$(4) a^{2} + 20 d^{2} = 120$
$(4) 5^{2} + 20 d^{2} = 120$
$d^{2} = 1$
$d = + 1 o r - 1$

$H e n c e n u m b e r s a r e 2, 4, 6, a n d 8$

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Four numbers are in A.P. If their sum is 20 and the sum of their squares is 120, then the middle terms are

The correct option is B 4,6 Let the numbers are a−3d,a−d,a+d,a+3d Given: a−3d+a—d+a+d+a+3d=20 4a=20 a=5 and (a−3d)2+(a−d)2+(a+3d)2+(a+3d)2=120 (4)a2+20d2=120 (4)52+20d2=120 d2=1 d=+1 or−1 Hence numbers are 2,4,6,and 8