Question

Which term of the

a.p. 5, 15, 25...will be 130 more than its 31 term?

Answer 1

heya......


First we need to find the 31st term

The formula for finding 31st term is a + 30d


So lets find 

An = a + (n-1)d= 5 +( 31-1)*(10)= 5 + 300 = 305

Therefore 31st term is 305!

Now 305 + 130 ( given in question) = 435


Now we need to find which term has number 435!


We need to find n

Therefore in this condition we have An = 435
So, 

An = a + (n-1)d435 = 5 + (n-1)10435-5 = 10n-10430+10=10n440 = 10n

Therefore 10n =440

Therefore n =44

It means answer is 44th term! 
Hope it helps! 


tysm....#gozmit

Answer 2

Hey mate!

Here's your answer!!

•Let nth term be 130 more than the 31st term of the A.P.

First term of A.P. = 5

Common difference = 15 – 5 = 10

an = 130 + a31

⇒ 5 + (n – 1) × 10 = 130 + 5 + (31 – 1) × 10

⇒ 10 (n – 1) = 430

⇒ n – 1 = 43

⇒ n = 44

Thus, 44th term of the A.P is 130 more than the 31st term.

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