Question

Which term of AP 5 15 25 and so on will be 130 more than its 31st term

Answer 1

Answer: the 44th term is 130 more than the 31st term

Step-by-step explanation:

Given AP = 5, 15, 25.....

Here, first term = a = 5 and common difference = d = 10

To find the “n”th term of an AP, we use the formula

$a_{n} = a + (n-1)*d$

Applying the formula and finding the 31st term

=> a(31st term) = 5 + 30*10 = 300 + 5 = 305

According to the question,

$a_{n} = 130 + a_{31}\\\\=> a_{n} = 130 + 305\\\\=> a + (n-1)*d = 435\\\\=> 5 + (n-1)*10 = 435\\\\=> (n-1)*10 = 430\\\\=> n-1 = 43\\\\=> n = 44$

Therefore, the 44th term of the AP is 130 more than the 31st term

Please brainlist my answer, if helpful!

Answer 2

a31=a+30d=5+30(10)=305
Acc’n to the Question
305+130=435;as 31st term+ 130
435=5+(n-1)10
435-5+1/10=430+1/10=43+1=44
.•. 44th term will be 130 more than its 31st term.
I hope this will help