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

In the sum of first n terms of an A.P. is cn2. then the sum of squares of these n terms is

A
n(4n21)c26
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
n(4n2+1)c23
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
n(4n21)c23
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
n(4n2+1)c26
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C n(4n21)c23
Given that for an A.P,Sn=cn2
Then Tn=SnSn1=cn2c(n1)2=(2n1)c
Sum of squares of n terms of this A.P.
=T2n==(2n1)2.c2=c2[4n24n+n]=c2[4n(n+1)(2n+1)64n(n+1)2+n]=c2n[2(2n2+3n+1)6(n+1)+33]=c2n[4n213]=n(4n21)c23

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Summation by Sigma Method
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon