Question

15. Which term of the AP 5, 15, 25, ... will be 130 more than its 31st term?
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Answer 1

Answer:

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