Question

which term of AP 5,15,25,.....will be 130 more than it's 31st term​

Answer 1

Answer:

44th term of the A.P will be 130 more than the first term.

Step-by-step explanation:

First we need to calculate 31st term -

A31 = a+(n-1)d

d (common difference) = 10, first term of A.P (a) = 5

then we get -> 5+(31-1) 10

31st term of the A.P = 305

     Now we will add 130 to our 31st term we get 435.

as we need to find 'n' we will keep it as a variable.

Substituting into equation :-

An = a+(n-1)d

435 = 5+(n-1)10

435 = 5+10n-10

435 = 10n-5

435+5 = 10n

440 = 10n

n = 44.

Hope this helps you.

Answer 2

${\large{\text{\underline{\underline{\red{Given :}}}}}}$

$\text{5,15,25 are in A.P}$

${\large{\text{\underline{\underline{\red{To Find :}}}}}}$

$\text{Which term of A.P is 130 more than it's 31 st term}$

${\large{\text{\underline{\underline{\red{Formula Applied :}}}}}}$

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

$n = (\frac{l-a}{d})+1$

${\large{\text{\underline{\underline{\red{Solution :}}}}}}$

$\text{It's given that 5,15,25 are in A.P hence,}$

$a=\text{first term}\implies \bf 5$

$d=\text{common difference}\implies t_2-t_1 = 15-5\implies \bf10$

$\text{Let's find the 31 st term of A.P}$

$n=\text{No.of.term}\implies \bf31$

$\implies t_n=a+(n-1)d$

$\implies t_{31}=5+(31-1)10$

$\implies t_{31}=5+(30)10$

$\implies t_{31} = 5+300$

$\implies t_{31} = 305$

$\text {Now let's find the term that is 130 more than its 31 st term}$

$\implies t_n= t_{31}+130$

$\implies t_n = 305+130$

$\implies t_n=435 , n= ?$

$n = (\frac{l-a}{d})+1$

$a=5,d=10,l=435,n= ?$

${\red{\text{Substitute the values in formula}}}$

$n =( \frac{435-5}{10})+1$

$n = (\frac{430}{10})+1$

$n = 43+1$

${\red{n = 44}}$

$\text{44 th term of A.P is 130 more than it's 31 st term}$

${\large{\text{\underline{\underline{\red{Verification :}}}}}}$

$\implies t_{44} = t_{31}+130$

$\implies t_{44} = 305+130 \implies 435$

$\implies t_{44} = a+(n-1)d$

$\implies t_{44} = 5+(44-1)10$

$\implies t_{44} = 5+(43)10$

$\implies t_{44} = 5+430$

$\implies t_{44} = 435$

$\text{Hence verified}$