Question

If $n$ A. M.s are inserted between $20$ and $80$ such that first mean: last mean =$1:3$.Find the value of $n$

• A. $13$
• B. $12$
• C. $10$
• D. $11$

Answer 1

The correct option is D $11$

$a=20$(first term) $l=80$(last\quad term)

no of $A{M}^{\prime }s=n$

$\therefore$total no of term in $AP=$no of $means+$one first term$+$one last term.

$N=n+2$

let the common difference $=d$

$l=a+(N-1)d80=20+(n+2-1)dd=\frac{60}{n+1}.........\left(1\right)$

Now, first $mean=a+d=20+d$

Last $mean=a+nd=20+ndRatio=1:3\therefore \frac{20+d}{20+nd}=\frac{1}{3}60+3d=20+nd40=d(n-3)d=\frac{40}{(n-3)}...............\left(2\right)$

Equating $\left(1\right)$and$\left(2\right)$,we get

$\frac{60}{(n+1)}=\frac{40}{(n-3)}\therefore 60n-180=40n+40\therefore 20n=220=>n=11$