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8. Sequences and Series
hard
If $n$ arithmetic means are inserted between a and $100$ such that the ratio of the first mean to the last mean is $1: 7$ and $a+n=33$, then the value of $n$ is
A
$21$
B
$22$
C
$23$
D
$24$
(JEE MAIN-2022)
Solution
$d =\frac{100- a }{ n +1}$
$A _{1}= a + d$
$A _{ n }=100- d$
$\Rightarrow \frac{ A _{1}}{ A _{ n }}=\frac{1}{7} \Rightarrow \frac{ a + d }{100- d }=\frac{1}{7}$
$\Rightarrow 7 a+8 d=100$
$\Rightarrow 7\, a +8\left(\frac{100- a }{ n +1}\right)=100$……..$(1)$
$\because a + n =33$………(2)
$Now,\,by\, Eq. (1) and (2)$
$7 n^{2}-132 n-667=0$
$n =23$ and $n =\frac{-29}{7}$ $reject.$
Standard 11
Mathematics
