Question

If 8th term of an AP is zero,prove that it's 38th term is triple of its 18th term

Answer 1

Hey There,


The solution is in the image.


Hope it helps,
Purva
Brainly Community

Answer 2

GIVEN:-

$\bf \: •8th \: term \: of \: AP \: is \: zero$

TO PROVE:-

$\bf \: 38th \: term \: of \: AP \: is \: triple \: its \: 18th \: term$

SOLUTION:-

$\bf In \: the \: given \: AP \: let \: a \: and \: d \: be \: the \: first \: and \: last \: term.$

$\bf Then,T_n=a+(n-1)d$

$\bf \implies T_8=a+(8-1)d,T_{38}=a+(38-1)d,T_{18}=a+(18-1)d$

$\bf \implies T_8=(a+7d),T_{38}=(a+37d),T_{18}=(a+17d)$

$\bf \: Now ,T_8=0\implies a+7d=0\implies a=-7d \: \: \: \: ------(1)$

$\bf \therefore \: T_{38}=(a+37d)=(-7d+37d)=30d \: \: \: [using (1)]$

$\bf And ,T_{18}=(a+17d)=(-7d+17d)=10d \: \: \: \: [using \: (1)]$

$\bf \therefore T_{38}=30d=3×(10d)=3×T_{18}$

$\bf \: Hence \: ,38th \: term \: is \: triple \: its \: 18th \: term$