Question

Which term of the ap 5,15,25......... Will be 130 more than its 31st terms?

Answer 1

Answer:

Step-by-step explanation:

Its very simple

In this ,

D(common diff.) = 10

A(first number) = 5

First lets take out 31st term , which is

= a +30d

= 5 + 30×10

= 305

Let the term which will have 130 more then its 31st term be X .

Hence X = 305+ 130

= 435

a + (n-1)d = 435

5 + (n-1)10 = 435

5 + 10n - 10 = 435

10n = 440

n = 440/10

n = 44

Hence the ans id 44th term

Answer 2

Answer:

44

Step-by-step explanation:

In this AP the first term = a = 5;

a1= 5;

a2= 15;

a3 = 25;

Common Difference = d = a2 - a1 = a3- a2  = 15 - 5 = 10;

The general Term aₙ = a + (n-1)d

                                  = 5 + (n-1)10

                                  = 5 + 10n -10

                                  = 10n -  5

Putting n=31; a₃₁ = 10(31) - 5 = 310 - 5 = 305

Let the nth term be 130 more than its 31 st term

aₙ = a₃₁ + 130

10n - 5 = 305 + 130

10n - 5 =435

10n = 440

n = 44

Therefore the 44th term is 130 more than the 31st term