Question

when n arithmetic means are inserted between 8 and 44 the sum of resulting term is 338. find the value of n and common difference

Answer 1

$\Huge\text{ n=11 , d=3 }$

$\Large\textrm{Given: -}$

"When $n$ arithmetic means are inserted between 8 and 44, the sum of the resulting term is 338. Find the value of $n$ and common difference."

$\Large\textrm{Let us denote: -}$

$\textrm{ a_{n} as the n th term}$

$\textrm{ S_{n} as the sum of the first n terms}$

$\Large\textrm{It is given that: -}$

$a_{1}=8$

$a_{n+2}=44$

$S_{n+2}=338$

$\textrm{\underline{ n is odd} since the median exists}$

$\Large\textrm{It follows that: -}$

$S_{n+2}=\dfrac{(n+2)(8+44)}{2}$

$S_{n+2}=26(n+2)$

$26(n+2)=338$

$n+2=13$

$n=11$

Hence the value of $n$ is 11.

To find the common difference, we use the first and the last term.

[tex]\begin{cases} & a_{1}=8 \\ & a_{13}=44 \end{cases}[/tex]

$\begin{cases} & a=8 \\ & a+12d=44 \end{cases}$

$d=3$

$\Large\textrm{Formulae used: -}$

$\large\textrm{ \bigstar Sum of A.P. \bigstar }$

$\cdots\longrightarrow\boxed{S_{n}=\dfrac{n(a+l)}{2}}$

If we substitute $l=a+(n-1)d$, we get the following.

 $\cdots\longrightarrow\boxed{S_{n}=\dfrac{n\{2a+(n-1)d\}}{2}}$

Each formula has different usage. If we know $a,l,n$ we use the first formula. If we know $a,d,l$ we use the second formula.