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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 12 of 106
Marks: +1, -0
a1a_1 = - 1 , ana_n = an1n\frac{a_n-1}{n} , n ≥ 2
Solution:  
We have given,
a1a_1 = - 1 , ana_n = an1n\frac{a_n-1}{n} , n ≥ 2
a1a_1 = - 1 , a2a_2 = a12\frac{a_1}{2} = 12-\frac{1}{2} , a3a_3 = a23\frac{a_2}{3} = 123\frac{-\frac{1}{2}}{3} = 16-\frac{1}{6}
a4a_4 = a34\frac{a_3}{4} = 164\frac{-\frac{1}{6}}{4} = 124-\frac{1}{24} , a5a_5 = a45\frac{a_4}{5} = 1245\frac{-\frac{1}{24}}{5} = 1120-\frac{1}{120}
Hence the first five terms of the given sequence are – 1, – 1/2, – 1/6, –1/24, –1/120.
The corresponding series is
- 1 + (12)+(16)\left(-\frac{1}{2}\right)+\left(-\frac{1}{6}\right) + (124)+(1120)\left(-\frac{1}{24}\right)+\left(-\frac{1}{120}\right) + ...
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