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  • Find the number of terms in the sequence <img alt="1,-1,\frac{-1}{3},\frac{-1}{5},\frac{-1}{7},......\frac{-1}{141}" src="https://entrancecorner.oncodecogs.com/gif.latex?1%2C-1%2C%5Cfrac%7B-1%

Find the number of terms in the sequence 1,-1,\frac{-1}{3},\frac{-1}{5},\frac{-1}{7},......\frac{-1}{141}

 

Option: 1

70


Option: 2

72


Option: 3

74


Option: 4

73


Answers (1)

best_answer

Clearly the given sequence 

1,-1,\frac{-1}{3},\frac{-1}{5},\frac{-1}{7},......\frac{-1}{141}  is an HP.

Let total number of terms be n.

Corresponding AP is 1,-1,-3......-141

It also has n terms.

So 

-141=1+(n-1)(-2)

-142=(n-1)(-2)

n-1=71

n=72

So, total 72 terms are there.

Posted by

Rakesh

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