# If n arithmetic means are interested between $0$ and $32$ such that the $2^{nd}$ mean is one-seventh of the $(n-1)^{th}$ mean, find the value of $n$.

$\begin{array}{1 1} 7 \\ 14 \\ 15 \\ 8 \end{array}$

## 1 Answer

Answer : (c) 15
$Explanation \; : \;n\;AMs\;between\;0\;\xi\;32$
$2^{nd}\;mean=a+2d$
$=0+2d=2d$
$(n-1)^{th}\;mean=\;32-2d$
$Given\;that\;\frac{2d}{32-2d}=\frac{1}{7}$
$14d=32-2d$
$16d=32$
$d=2$
$Since\;the\;AP\;has\;its\;(n+2)^th\;term\;as\;32$
$32=a+(n+2-1)d$
$=0+(n+1)2$
$16=n+1$
$n=15.$
answered Jan 3, 2014 by

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