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

Consider the real-valued function $h: \{0, 1, 2, \ldots, 100\} \rightarrow \mathbb{R}$ such that $h(0) = 5$,$h(100) = 20$,and satisfying $h(p) = \frac{1}{2}\{h(p+1) + h(p-1)\}$ for every $p = 1, 2, \ldots, 99$. Then the value of $h(1)$ is:

Let the sum of $n, 2n, 3n$ terms of an $A.P.$ be $S_{1}, S_{2}$ and $S_{3}$ respectively. Show that $S_{3} = 3(S_{2} - S_{1})$.

Let the sum of the first three terms of an $A.P.$ be $39$ and the sum of its last four terms be $178.$ If the first term of this $A.P.$ is $10,$ then the median of the $A.P.$ is

If $a, b, c$ are positive real numbers such that $ab^2c^3 = 64$,what is the minimum value of $(1/a + 2/b + 3/c)$?

Find the sum of all natural numbers lying between $100$ and $1000$ which are multiples of $5.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module