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NCERT Exemplar · Q25

Q.If tnt_n denotes the nnth term of the series 2+3+6+11+18+…2 + 3 + 6 + 11 + 18 + \ldots then t50t_{50} is
(A) 492−149^2 - 1
(B) 49249^2
(C) 502+150^2 + 1
(D) 492+249^2 + 2

Sikkim CbseMCQ· 1mImportance★★★★★est
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The successive differences are 1,3,5,7,…1, 3, 5, 7, \ldots (odd numbers), giving tn=2+(n−1)2t_n = 2 + (n-1)^2, so t50=492+2t_{50} = 49^2 + 2 — option (D).

For the series 2+3+6+11+18+…2 + 3 + 6 + 11 + 18 + \ldots, the first differences are

3−2=1,6−3=3,11−6=5,18−11=7,3-2=1,\quad 6-3=3,\quad 11-6=5,\quad 18-11=7,

i.e. the consecutive odd numbers 1,3,5,7,…1, 3, 5, 7, \ldots

Building up the nnth term from the first:

tn=t1+∑k=1n−1(2k−1)=2+(n−1)2,t_n = t_1 + \sum_{k=1}^{n-1}(2k-1) = 2 + (n-1)^2, …

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