An important and exam oriented question from arithmetic progression chapter as it is already asked in previous year paper of 2006 in which we are to find which term of the AP 5,15,25,….will be 130 more than its 31st term?
Book – RS AGGARWAL, class 10, Arithmetic Progression, chapter 5A, question no 15
We have, a=5 and d=10
an=a+(n−1)d
∴a31=a+30d=5+30×10=305
Let nth term of the given A.P. be 130 more than its 31st term.
Then, an=130+a31
∴a+(n−1)d=130+305
⟹5+10(n−1)=435
⟹10(n−1)=430
⟹n−1=43
⟹n=44
Hence, 44th term of the given A.P. is 130 more than its 31st term.