If n arithmetic means are inserted between a | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$d =\frac{100- a }{ n +1}$

$A _{1}= a + d$

$A _{ n }=100- d$

$\Rightarrow \frac{ A _{1}}{ A _{ n }}=\frac{1}{7} \Rightarrow \frac{ a + d }{10

Vedclass · Accepted Answer

$23$

Vedclass · Answer

$d =\frac{100- a }{ n +1}$

$A _{1}= a + d$

$A _{ n }=100- d$

$\Rightarrow \frac{ A _{1}}{ A _{ n }}=\frac{1}{7} \Rightarrow \frac{ a + d }{100- d }=\frac{1}{7}$

$\Rightarrow 7 a+8 d=100$

$\Rightarrow 7\, a +8\left(\frac{100- a }{ n +1}\right)=100$........$(1)$

$\because a + n =33$.........(2)

$Now,\,by\, Eq. (1) and (2)$

$7 n^{2}-132 n-667=0$

$n =23$ and $n =\frac{-29}{7}$ $reject.$

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

Similar Questions

Insert five numbers between $8$ and $26$ such that resulting sequence is an $A.P.$

The sequence $\frac{5}{{\sqrt 7 }}$, $\frac{6}{{\sqrt 7 }}$, $\sqrt 7 $, ....... is

If $A$ be an arithmetic mean between two numbers and $S$ be the sum of $n$ arithmetic means between the same numbers, then

Let $\frac{1}{{{x_1}}},\frac{1}{{{x_2}}},\frac{1}{{{x_3}}},.....,$ $({x_i} \ne \,0\,for\,\,i\, = 1,2,....,n)$ be in $A.P.$ such that $x_1 = 4$ and $x_{21} = 20.$ If $n$ is the least positive integer for which $x_n > 50,$ then $\sum\limits_{i = 1}^n {\left( {\frac{1}{{{x_i}}}} \right)} $ is equal to.

The sum of the numbers between $100$ and $1000$, which is divisible by $9$ will be