1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 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 __.

A

2,4

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B

4,6

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

6,8

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D

8,10

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is A 4,6Let the numbers be a−3d,a−d,a+d and a+3d. These numbers form an AP with first term a−3d and common difference d>0. Also, given that (a−3d)+(a−d)+(a+d)+(a+3d)=20. ⇒4a=20 ⇒a=5 Also, (a−3d)2+(a−d)2+(a+d)2+(a+3d)2=120(a2−6ad+9d2)+(a2−2ad+d2)+(a2+2ad+d2)+(a2+6ad+9d2)=120 ⇒4a2+20d2=120 ⇒4(5)2+20d2=120⇒4(25)+20d2=120 ⇒d2=1 ⇒d=±1 SIince d>0, we must have d=1.1st number = a−3d=5−3×1=22nd number = a−d=5−1=43rd number = a+d=5+1=64th number = a+3d=5+3×1=8 Hence, the numbers are 2, 4, 6, and 8.

Suggest Corrections
2
Join BYJU'S Learning Program
Related Videos
Formula for Sum of N Terms of an AP
MATHEMATICS
Watch in App
Join BYJU'S Learning Program