Insert 6 arithmetic means between 20 and 76. | vedclass

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

Question

(C) Let the six arithmetic means be $A_1, A_2, A_3, A_4, A_5, A_6$ between $20$ and $76$.Then,the sequence $20, A_1, A_2, A_3, A_4, A_5, A_6, 76$ form

Vedclass · Accepted Answer

$28, 36, 44, 52, 60, 68$

Vedclass · Answer

(C) Let the six arithmetic means be $A_1, A_2, A_3, A_4, A_5, A_6$ between $20$ and $76$.Then,the sequence $20, A_1, A_2, A_3, A_4, A_5, A_6, 76$ forms an arithmetic progression $(AP)$ with $n = 8$ terms.The first term $a = 20$ and the last term $a_8 = 76$.The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.Substituting the values,$76 = 20 + (8 - 1)d$.$76 - 20 = 7d \Rightarrow 56 = 7d \Rightarrow d = 8$.The arithmetic means are:$A_1 = 20 + 8 = 28$$A_2 = 28 + 8 = 36$$A_3 = 36 + 8 = 44$$A_4 = 44 + 8 = 52$$A_5 = 52 + 8 = 60$$A_6 = 60 + 8 = 68$Thus,the arithmetic means are $28, 36, 44, 52, 60, 68$.

Insert $6$ arithmetic means between $20$ and $76$.

Similar Questions

The sum,$\sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4}$ is equal to

The sum of the first $20$ terms of the sequence $0.7, 0.77, 0.777, \dots$ is

The sum of all terms of the $n^{th}$ bracket of the sequence $(1), (3, 5), (7, 9, 11), \dots$ is equal to:

If $a, b, c$ are in $G.P.$ and $\log a - \log 2b, \log 2b - \log 3c$ and $\log 3c - \log a$ are in $A.P.$,then $a, b, c$ are the lengths of the sides of a triangle which is

Let $n (> 1)$ be a positive integer. Then,the largest integer $m$ such that $(n^m + 1)$ divides $(1 + n + n^2 + \dots + n^{127})$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module