Question

The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

Answer 1

We know that

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

Given, $T_4=a+(4-1)d=0$

$a+3d=0$

$a=-3d$

$T_{25}=a+(25-1)d$

= $a+24d=(-3d)+24d$

= $21d$

And, $T_{11}=a+(11-1)d$

= $a+10d$

Then, $3T_{11}=3(a+10d)$

= $3a+30d$

= $3(-3d)+30d(a=-3d)$

= $30d-9d=21d=T_{25}$

$\therefore 3T_{11}=T_{25}$ Hence Proved.