Sum the series 51 + 50 + 49 + ldots + 21. | vedclass

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

Question

(A) The given series is an arithmetic progression where the first term $a = 51$,the common difference $d = -1$,and the last term $a_n = 21$.Using the

Vedclass · Accepted Answer

$1116$

Vedclass · Answer

(A) The given series is an arithmetic progression where the first term $a = 51$,the common difference $d = -1$,and the last term $a_n = 21$.Using the formula for the $n^{th}$ term of an arithmetic progression: $a_n = a + (n - 1)d$.Substituting the values: $21 = 51 + (n - 1)(-1)$.$21 = 51 - n + 1$.$21 = 52 - n$.$n = 52 - 21 = 31$.Now,using the sum formula $S_n = \frac{n}{2}(a + a_n)$:$S_{31} = \frac{31}{2}(51 + 21)$.$S_{31} = \frac{31}{2}(72)$.$S_{31} = 31 	imes 36 = 1116$.

Sum the series $51 + 50 + 49 + \ldots + 21$.

Similar Questions

The sum of a series in $A.P.$ is $525$. Its first term is $3$ and last term is $39$. Find the common difference.

The first term of an $A.P.$ is $2$ and the common difference is $4$. The sum of its $40$ terms will be:

For a series $S = 1 - 2 + 3 - 4 + \dots$ up to $n$ terms,
Statement $-1$: The sum of the series is always dependent on the value of $n$,i.e.,whether it is even or odd.
Statement $-2$: The sum of the series is $-\frac{n}{2}$ when $n$ is any even integer.

Let $S_n$ denote the sum of the first $n$ terms of an $A.P$. If $S_4 = 16$ and $S_6 = -48$,then $S_{10}$ is equal to

The sum of all the digits of the numbers from $1$ to $100$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module