Question

The $4^{th}$ term of $AP$ is $0$ prove that ${25}^{th}$ term of $AP$ is $3$ times $a_{11}$ term ?

Answer 1

$a_4=a+(4-1)d=a+3d=0$

$a=-3d$

$a_{25}=a+(25-1)d=a+24d=-3d+24d=21d$

$a_{11}=a+(11-1)d=a+10d=-3d+10d=7d$

$\frac{a_{25}}{a_{11}}=\frac{21}{7}=3a_{25}=3a_{11}$