Class 11

Math

Algebra

Sequences and series

There are $n$ Arithmetic means between $13$ and $61.$ If ratio of the first $A.M$. and $(n−1)$ the $A.M.$ may be $16:55,$ then find $n.$

Then, the common difference is given by:

$d=n+1b−a $

$d=n+161−13 $

$d=n+148 $

Now,

$A_{1}=a+d=13+n+148 $

$A_{1}=n+113n+61 $

And

$A_{n}=a+nd=13+n+148n $

$A_{n}=n+161n+13 $

Since,

$A_{n}A_{1} =5516 $

$61n+1313n+61 =5516 $

$715n+3355=976n+208$

$261n=3147$

$n≈12$

Hence, this is the answer.