Question

Which term of an AP 5,15,25,.....Will have 130 morethan its 31st term?

Answer 1

a=5,d=10
31st term =a+30d
=5+30*10=5+300
=305
Given 5,15,25…..will be 130 more than 31st term.
Therefore,
=130+305=435
Now a+(n-1)d
435=5+(n-1)10
435=5+10n-10
435=-5+10n
435+5=10n
440=10n
n=440/10
=44th term

Answer 2

Given A.P. 5, 15, 25.....

Here, first term, a = 5

Common difference, d = 15 - 5 = 10

We have to find the term which will have a valua of 130 more than the $31^{st}$ Term.

First, we will find the $31^{st}$ Term.

We know that,

$T_n$ = a + (n - 1)d

So, $31^{st}$ Term =

5 + (31 - 1)10

→ 5 + (30)10

→ 5 + 300

= 305

Now, the number having 130 more value than 305 = 130 + 305 = 435

So we have to find the term which has a value of 435.

Let the term be n, so we get

So,

435 = a + (n - 1)d

→ 435 = 5 + (n - 1)10

→ 435 -5 = 10n - 10

→ 430 + 10 = 10n

→ n = 440/10

→ n = 44

So, the $44^{th}$ term of given A.P. would have 130 more value than the $31^{st}$ term.