The four arithmetic means between 3 and 23 ar | vedclass

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

Question

(B) Let the four arithmetic means be $A_1, A_2, A_3,$ and $A_4$.Then,the sequence $3, A_1, A_2, A_3, A_4, 23$ forms an Arithmetic Progression $(AP)$.H

Vedclass · Accepted Answer

$7, 11, 15, 19$

Vedclass · Answer

(B) Let the four arithmetic means be $A_1, A_2, A_3,$ and $A_4$.Then,the sequence $3, A_1, A_2, A_3, A_4, 23$ forms an Arithmetic Progression $(AP)$.Here,the first term $a = 3$ and the sixth term $T_6 = 23$.The formula for the $n^{th}$ term of an $AP$ is $T_n = a + (n-1)d$.For $n = 6$,we have $23 = 3 + (6-1)d$.$23 = 3 + 5d$$20 = 5d$$d = 4$.Now,we calculate the means:$A_1 = a + d = 3 + 4 = 7$$A_2 = a + 2d = 3 + 8 = 11$$A_3 = a + 3d = 3 + 12 = 15$$A_4 = a + 4d = 3 + 16 = 19$Thus,the four arithmetic means are $7, 11, 15, 19$.

The four arithmetic means between $3$ and $23$ are

Similar Questions

The sum of the series $3 + 33 + 333 + \dots + n$ terms is

If the $A.M.$ between $p^{th}$ and $q^{th}$ terms of an $A.P.$ is equal to the $A.M.$ between $r^{th}$ and $s^{th}$ terms of the same $A.P.$,then $p + q$ is equal to

The solution of the equation $(x + 1) + (x + 4) + (x + 7) + \dots + (x + 28) = 155$ is

If $S_n = \frac{n(n + 1)(n + 2)}{6}$,then $\sum_{n = 1}^\infty \frac{1}{t_n} = $

In a geometric progression,if the ratio of the sum of first $5$ terms to the sum of their reciprocals is $49$,and the sum of the first and the third term is $35$. Then the first term of this geometric progression is

Vedclass Test Series

Exam Paper Generator

Online Exam Module