After inserting n arithmetic means (A.M.s) be | vedclass

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

Question

(B) The resulting progression consists of $n + 2$ terms,where the first term $a = 2$ and the last term $l = 38$.The sum of an arithmetic progression w

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The resulting progression consists of $n + 2$ terms,where the first term $a = 2$ and the last term $l = 38$.The sum of an arithmetic progression with $N$ terms is given by $S_N = \frac{N}{2}(a + l)$.Here,$N = n + 2$,$a = 2$,and $l = 38$.So,the sum is $S = \frac{n + 2}{2}(2 + 38) = \frac{n + 2}{2}(40) = 20(n + 2)$.Given that the sum is $200$,we have $20(n + 2) = 200$.Dividing by $20$,we get $n + 2 = 10$.Therefore,$n = 8$.

After inserting $n$ arithmetic means $(A.M.s)$ between $2$ and $38$,the sum of the resulting progression is $200$. The value of $n$ is

Similar Questions

If the ratio of the sum of $n$ terms of two $A.P.s$ is $(7n + 1):(4n + 27)$,then the ratio of their $11^{th}$ terms is:

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

If the $m^{th}$ terms of the series $63 + 65 + 67 + 69 + \dots$ and $3 + 10 + 17 + 24 + \dots$ are equal,then $m = $

The sum of all two-digit numbers which,when divided by $4$,yield unity as a remainder is:

If $\log 2$,$\log (2^x - 1)$,and $\log (2^x + 3)$ are in arithmetic progression,then the value of $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module