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JEE Main 11 Jan 2019 Shift 1 Solved Paper
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© examsnet.com
Question : 34
Total: 90
The sum of an infinite geometric series withpositive terms is 3 and the sum of the cubes ofits terms is
27
19
. Then the common ratio of thisseries is :
4
9
2
9
2
3
1
3
Validate
Solution:
a
1
−
r
= 3 ... (1)
a
3
1
−
r
3
=
27
19
⇒
27
(
1
−
r
)
3
1
−
r
3
=
27
19
⇒
6
r
2
- 13r + 6 = 0
⇒ r =
2
3
as |r| < 1
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