JEE Main 2022 (Online) 28th June Evening ShiftQuestion 1
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 :
Sequence and Series · Means and Inequalities (AM, GM, HM)
Solution
Formulas used. (1) $n$ A.M.s between $a$ and $b$: common difference $d=\frac{b-a}{n+1}$(2) First mean $=a+d$, last mean $=a+nd$Step 1 - Ratio condition. $$\frac{a+d}{a+nd}=\frac17\ \Longrightarrow\ 7a+7d=a+nd\ \Longrightarrow\ 6a=(n-7)d .$$ Step 2 - Put $d=\dfrac{100-a}{n+1}$. $$6a(n+1)=(n-7)(100-a).$$ Step 3 - Put $a=33-n$. Then $100-a=67+n$: $$6(33-n)(n+1)=(n-7)(n+67)\ \Longrightarrow\ -6n^2+192n+198=n^2+60n-469 .$$ $$7n^2-132n-667=0\ \Longrightarrow\ n=\frac{132\pm190}{14}.$$ $n$ is positive, so $n=\dfrac{322}{14}=23$. Check. $a=10$, $d=\dfrac{90}{24}=\dfrac{15}{4}$; first mean $\dfrac{55}{4}$, last mean $10+23\cdot\dfrac{15}4=\dfrac{385}{4}$, ratio $1:7$. Answer: (C)