wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

3,1(3),13(3).....
The 19th term of the above sequence is .

A
3318
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
3152
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
37326
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 3318
In the given sequence,
t2t1=133=13.
Common ratio, r=13
and
first term, a=3
Now, a19=ar191.
=3×(13)18=3318

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
General Term of an GP
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon