Question

Q9 : Which term of the A.P 5, 15, 25, . . . . . . will be 130 more than its 31st term ?

1 point

44

40

54

none of these

​

Answer 1

$\bold\blue{Question}$

$\bold{Which \ term \ of \ A.P. \ 5,15,25,... \ will}$

$\bold{be \ 130 \ more \ than \ its \ 31^{st} \ term?}$

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{44^{th} \ term \ is \ 130 \ more \ than \ 31^{st} \ term.}$

$\bold\orange{Given:}$

$\bold{=>A.P. \ 5,15,25,...}$

$\bold\pink{To \ find:}$

$\bold{The \ term \ which \ is \ 130 \ more \ than \ 31^{st} \ term.}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{The \ given \ A.P. \ is}$

$\bold{=>5,15,25,...}$

$\bold{Here, \ a=5, \ d=15-5=10}$

$\bold{tn=a+(n-1)d...formula}$

$\bold{t31=5+(31-1)×10}$

$\bold{\therefore{t31=5+30×10}}$

$\bold{\therefore{t31=305}}$

__________________________________

$\bold{Let \ tn \ be \ 130 \ more \ than \ 31^{st} \ term.}$

$\bold{\therefore{tn=t31+130}}$

$\bold{tn=305+130}$

$\bold{tn=435}$

$\bold{tn=a+(n-1)d...formula}$

$\bold{435=5+(n-1)×10}$

$\bold{\therefore{(n-1)×10=435-5}}$

$\bold{\therefore{(n-1)×10=430}}$

$\bold{\therefore{n-1=\frac{430}{10}}}$

$\bold{\therefore{n-1=43}}$

$\bold{\therefore{n=43+1}}$

$\bold{\therefore{n=44}}$

$\bold\purple{\tt{\therefore{44^{th} \ term \ is \ 130 \ more \ than \ 31^{st} \ term.}}}$