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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 41 of 106
Marks: +1, -0
0.15, 0.015, 0.0015, ... 20 terms.
Solution:  
In the given G.P.
a = 0.15, r = 0.1, n = 20
∴ SnS_n = a(1−rn)1−r\frac{a(1-r^n)}{1-r} , if |r| < 1
S20S_{20} = a(1−r20)1−r\frac{a(1-r^{20})}{1-r} = (0.15) 1−(0.1)201−0.1\frac{1-(0.1)^{20}}{1-0.1} = 0.150.9[1−(0.1)20]\frac{0.15}{0.9}[1-(0.1)^{20}] = 16[1−(0.1)20]\frac{1}{6}[1-(0.1)^{20}]
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