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.

Share

Become a Mathematics wizard and get above

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

a= 3dÂ =Âa_{2}Â âˆ’Âa_{1}Â = 15 âˆ’ 3 = 12We know:-

aÂ =Â_{n}a+(nâˆ’1)Therefore,

a

_{54}Â =Âa+(54âˆ’1)d3+(53)(12)

3+636 = 639

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

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

aÂ =Â_{n}a+(nâˆ’1)d771 = 3+(

nÂ âˆ’1)12768 = (

nâˆ’1)12(

nÂ âˆ’1) = 64nÂ = 65Hence, the 65 term was 132 more than 54th term.