Question

$n$ arithmetic means are inserted between $3$ and $17$. If the ratio of last and the first arithmetic mean is $3:1$, then the value of $n$ is

• A. $9$
• B. $6$
• C. $7$
• D. $5$

Answer 1

Answer: (B)

$6$

$3,a_1,a_2,....a_n,17ϵA.P.(n+2terms)$
$\Rightarrow 3,3+d,a_2,a_2.....17-d,17ϵA.P.$
$\Rightarrow \frac{17-d}{3+d}=\frac{3}{1}$
$\therefore 17-d=9+3d\Rightarrow d=2$
Now $a+(n+1)d=t_{n+2}=17$
$\Rightarrow 3+(n+1)\left(2\right)=17\Rightarrow n=6$
Hence choice (b) is correct.