Question

rms.
10. Which term of the A.P. 5, 15, 25, ...... will be 130
more than its 31st term?​

Answer 1

Question : -

• Which term of the A.P. 5, 15, 25, ...... will be 130 more than its 31st term?

Answer

Given :-

• An AP series 5, 15, 25, .....

To find :-

• Term which is 130 more than 31st term.

Formula used :-

nth term of an AP series is given by

$\bf \:a_n = a + (n - 1) \times d$

where,

• a = first term of an AP
• d = Common Difference of an AP

Solution :-

The AP series is 5, 15, 25, ......

First term, a = 5

Common Difference, d = 15 - 5 = 10.

Let nth term of an AP is 130 more than 31st term.

$\bf\implies \:a_n = 130 + a_{31}$

$\bf\implies \:a + (n - 1) \times d = 130 + a + 30d$

Cancel 'a' and put the value of d, we get

$\bf\implies \:(n - 1) \times 10 = 130 + 30 \times 10$

$\bf\implies \:(n - 1) \times 10 = 430$

$\bf\implies \:n - 1 = 43$

$\bf\implies \:n = 44$

Hence, 44th term is 130 more than 31st term.