If sum_{r=1}^n(2r+1)=440,then n = ldots. | vedclass

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

Question

(A) Given the sum $\sum_{r=1}^n(2r+1) = 440$. Expanding the summation,we get the series $3 + 5 + 7 + \ldots + (2n+1) = 440$. This is an arithmetic pro

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) Given the sum $\sum_{r=1}^n(2r+1) = 440$. Expanding the summation,we get the series $3 + 5 + 7 + \ldots + (2n+1) = 440$. This is an arithmetic progression with first term $a = 3$,common difference $d = 2$,and number of terms $n$. The sum of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n-1)d]$. Substituting the values: $\frac{n}{2}[2(3) + (n-1)(2)] = 440$. $\Rightarrow \frac{n}{2}[6 + 2n - 2] = 440$. $\Rightarrow \frac{n}{2}[2n + 4] = 440$. $\Rightarrow n(n + 2) = 440$. $\Rightarrow n^2 + 2n - 440 = 0$. Solving the quadratic equation: $(n + 22)(n - 20) = 0$. Since $n$ must be positive,$n = 20$.

If $\sum_{r=1}^n(2r+1)=440$,then $n = \ldots$.

Similar Questions

Let $T_r$ be the $r^{th}$ term of an $A.P.$ for $r = 1, 2, 3, \dots$. If for some positive integers $m, n$ we have $T_m = \frac{1}{n}$ and $T_n = \frac{1}{m}$,then $T_{mn}$ equals

If the sum of $n$ terms of an $A.P.$ is $(pn + qn^2),$ where $p$ and $q$ are constants,find the common difference.

How many terms of the $A.P.$ $-6, -\frac{11}{2}, -5, \ldots$ are needed to give the sum $-25$?

If the first term of an $A.P.$ is $10$,the last term is $50$,and the sum of all the terms is $300$,then the number of terms is:

The sum of the first and third term of an arithmetic progression is $12$ and the product of the first and second term is $24$. Find the first term.

Vedclass Test Series

Exam Paper Generator

Online Exam Module