Question

If $4A{M}^{\prime }s$ are inserted between $\frac{1}{2}$ and $3$ then third $AM$ is

• A. $-2$
• B. $2$
• C. $-1$
• D. $1$

Answer 1

Answer: (B)

$2$

Given,

$4A{M}^{\prime }s$ are insented between $\frac{1}{2}$ and $3$

$\Rightarrow \frac{1}{2},A_1,A_2,A_3,A_4,3$ with common difference $d$

$n^{th}$ term $t_n=a+(n-1)d$

$a=$ First term

$d=$ Common difference

$n=$ number of terms

$t_n=n^{th}$ term

$\Rightarrow 3=\frac{1}{2}+5d\Rightarrow 5d=3\frac{-1}{2}=\frac{5}{2}$

$\Rightarrow d=\frac{1}{2}$

Then $3rdAM=A_3=\frac{1}{2}+3d=\frac{1}{2}+3.\frac{1}{2}=\frac{4}{2}=2$

$\therefore 3^{rd}AM=2$