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If $\mathrm{n}$ arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is $1: 7$ and $\mathrm{a}+\mathrm{n}=33$ , then the value of $\mathrm{n}$ is :
Option: 1
$21$

Option: 2
$22$

Option: 3
$23$

Option: 4
$24$

Answers (1)

best_answer

Let common difference of the AP obtained be d.

$\mathrm{\therefore\: First\: A M=a+d}\\$

$\mathrm{Last \: mean =100-d}\\$

$\mathrm{Their \: ratio =\frac{a+d}{100-d}=\frac{1}{7} \Rightarrow 7 a+8 d=100}$ ..........(i)

$\text { Also } \mathrm{100=a+(n+2-1) d} \\$

$\mathrm{\Rightarrow 100=a+(n+1) d} \\$

$\mathrm{\Rightarrow 100=33-n+(n+1) d \quad(\text { given } a+n=33)} \\$

$\mathrm{\Rightarrow d=\frac{n+67}{n+1} }$

Using (i)

$\mathrm{ 7(33-n)+\frac{8(n+67)}{n+1}=100} \\$

$\mathrm{\Rightarrow 7 n^{2}-132 n-667=0} \\$

$\mathrm{\Rightarrow n=23 }$

Hence the correct option is 3.

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