Which term of the A.P. 3, 15, 27, 39… will be 132 more than its 54th term?

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The question is given from ncert Book of class 10th Ch. no. 5 Ex. 5.2 Q. 11. In the following question you have to find the term which will be 132 more than the 54th term of the given A.P. Give the solution.

Solution:-It is given in this A.P. that 3, 15, 27, 39, …

a= 3d=a_{2}−a_{1}= 15 − 3 = 12We know:-

a=_{n}a+(n−1)Therefore,

a

_{54}=a+(54−1)d3+(53)(12)

3+636 = 639

a_{54 }= 639We have to find the term of this series or A.P which is 132 more than a

_{54 }Therefore 771.Let the nth term be 771.

a=_{n}a+(n−1)d771 = 3+(

n−1)12768 = (

n−1)12(

n−1) = 64n= 65Hence, the 65 term was 132 more than 54th term.