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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 4 of 106
Marks: +1, -0
ana_n = 2n−36\frac{2n-3}{6}
Solution:  
We have, ana_n = 2n−36\frac{2n-3}{6}
Substituting n = 1, 2, 3, 4, 5, we get
a1a_1 = 2⋅1−36\frac{2\cdot 1-3}{6} = 2−36\frac{2-3}{6} = - 16\frac{1}{6} , a2a_2 = 2⋅2−36\frac{2\cdot 2-3}{6} = 4−36\frac{4-3}{6} = 16\frac{1}{6}
a3a_3 = 2⋅3−36\frac{2\cdot 3-3}{6} = 6−36\frac{6-3}{6} = 36\frac{3}{6} = 12\frac{1}{2} , a4a_4 = 2⋅4−36\frac{2\cdot 4-3}{6} = 8−36\frac{8-3}{6} = 56\frac{5}{6}
a5a_5 = 2⋅5−36\frac{2\cdot 5-3}{6} = 10−36\frac{10-3}{6} = 76\frac{7}{6}
∴ The first five terms are - 16,16,12,56,76\frac{1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}
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