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

2,4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4,6

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6,8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8,10

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
B
4,6

D
8,10

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
${(a - 3 d)}^{2} + {(a - d)}^{2} + {(a + d)}^{2} + {(a + 3 d)}^{2} = 120$

$a^{2} + 9 d^{2} - 6 a d + a^{2} + d^{2} - 2 a d + a^{2} + d^{2} + 2 a d + a^{2} + 9 d^{2} + 6 a d$

$(4) a^{2} + 20 d^{2} = 120$

$(4) 5^{2} + 20 d^{2} = 120$

$100 + 20 d^{2} = 120$

$d^{2} = 1$

$d = + 1 o r - 1$

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

Suggest Corrections

Similar questions

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

Q. If the sum of four consecutive terms of an AP is 20 and the sum of their squares is 120, then the numbers are .

The sum of four consecutive numbers in an A.P. with common difference $d > 0$ is 20. If the sum of their squares is 120, then the middle terms are __ and __.

Q. Four number are in arithmetic progression. If their sum is

20

and sum of their square is

120

, then find the number.

Q. If four numbers are in A.P such that their sum is 20 and sum of their squares is 120, them the numbers are

Join BYJU'S Learning Program

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

The correct options are B 4,6 D 8,10 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+d)2+(a+3d)2=120 a2+9d2−6ad+a2+d2−2ad+a2+d2+2ad+a2+9d2+6ad (4)a2+20d2=120 (4)52+20d2=120 100+20d2=120 d2=1 d=+1 or−1 Hence numbers are 2,4,6,and 8