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Number of terms in the sequence 1, 3, 6, 10, 15, …, 5050 is
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a
50
b
75
c
100
d
125
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Correct option is C
Let,S=1+3+6+10+15+,…,+tn (1)then S=1+3+6+10+,…,+tn−1+tn (2)(1)−(2)⇒0=(1+2+3+4+… to n terms )−tn⇒ tn=n(n+1)2Given, 5050=n(n+1)2⇒n2+n−10100=0⇒ n=−1±1+404002=−1±404012=−1±2012=−101,100∴ n=100. (∵ n is a positive integer)Similar Questions
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