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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 42
Total: 106
√7,√21,3√7 , ... n terms.
Solution:  
In the given G.P.
a = √7 , r = √3 , sn =
a(rn−1)
r−1
 , if |r| > 1
∴ Sn = √7
[(√3)n−1]
√3−1
 =
√7
√3−1
×
√3+1
√3+1
[(√3)n−1]
=
√7(√3+1)
3−1
[(√3)n−1]
∴ Sn =
√7
2
(√3+1)(3
n
2
−1)
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