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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 6 of 106
Marks: +1, -0
ana_n = nn2+54n\frac{n^2+5}{4}
Solution:  
We have, ana_n = nn2+54n\frac{n^2+5}{4}
Substituting n = 1, 2, 3, 4, 5, we get
a1a_1 = 1.12+541.\frac{1^2+5}{4} = 64\frac{6}{4} = 32\frac{3}{2} , a2a_2 = 2.22+542.\frac{2^2+5}{4} = 2.94\frac{2.9}{4} = 92\frac{9}{2}
a3a_3 = 3.35+543.\frac{3^5+5}{4} = 3.144\frac{3.14}{4} = 212\frac{21}{2} , a4a_4 = 4.42+544.\frac{4^2+5}{4} = 4.214\frac{4.21}{4} = 21
a5a_5 = 5.55+545.\frac{5^5+5}{4} = 5.304\frac{5.30}{4} = 752\frac{75}{2}
∴ The first five terms are 32,92,212,21,752\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}
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