Question

If 4 times the fourth term of an AP is equal to 11 times its eleventh term, then its fifteenth term is equal to:

[1 Mark]

• A. 15
• B. 14
• C. 11
• D. 0

Answer 1

The correct option is D 0

Given,
$4\times t_4=11\times t_{11}$
$\Rightarrow 4(a+3d)=11(a+10d)$
[using $t_n=a+(n-1)d]$
$\Rightarrow 4a+12d=11a+110d$
$\Rightarrow 7a+98d=0$
$\Rightarrow a+14d=0$
$\Rightarrow a+(15-1)d=0$
$\Rightarrow t_{15}=0$