Question

which term of an AP 5,15,25.... will be 130 more than its 31st term

Answer 1

Ap is 5,15,25.......
a=5 and d=15-5=10
31st term=a+30d
=5+30*10
=5+300
31st term =305
an=305+130
=435
a+(n-1)d=435
5+(n-1)10=435
(n-1)10=435-5
(n-1)=430/10
n-1=43
n=44
It means the 44th term of the given AP will be 130 more than its 31st term.

Answer 2

Given :

An AP series

To Find :

Which term of an AP 5,15,25.... will be 130 more than its 31st term

Solution :

The nth term of an AP series can be given as -

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

where, $a_n$ = nth term; a = First term; d = common difference

In this question, a=5 and d=10

So, 31st term (a₃₁)  =   a+30d

                              =   5+30×10

                              =   5+300

                              =   305

The value of the term which is 130 more than the 31st term = 305 + 130

                                                                                                  =  435

As per question,

⇒   435     =  5+(n-1)10

⇒   435     =  5+10n-10

⇒   435     =  -5+10n

⇒  435+5  =  10n

⇒   440     =   10n

∴      n        =  44

Therefore, 44th term is 130 more than the 31st term in the given AP series.