Question

What is the sum of the following 41 counting numbers?
10 + 11 + 12 + ... + 50 =-

Answer 1

1230

Explanation:

Sum of integers from 1 to n:

(n)(n + 1)/2

Find the sum of integers from 1 to 50 and subtract the sum of integers from 1 to 9.

(50)(51)/2 - (9)(10)/2 = 1230

Answer 2

1230

Explanation:

Counting numbers, also known as natural numbers, are the set of positive integers starting from 1 and continuing indefinitely.

To find the sum of 41 counting numbers starting from 10, we can use sum of an arithmetic series formula.

`\boxed{\begin{minipage}{7.3 cm}\underline{Sum of the first $n$ terms of an arithmetic series}\\\\$S_n=(n)/(2)[2a+(n-1)d]$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the first term. \\ \phantom{ww}$\bullet$ $d$ is the common difference.\\ \phantom{ww}$\bullet$ $n$ is the position of the term.\\\end{minipage}}`

We want to find the sum of 41 counting numbers, so n = 41.

The first term is 10, so a = 10.

The common difference between consecutive terms is 1, so d = 1.

Substitute these values into the formula and solve.

`\begin{aligned}S_(41)&=(41)/(2)\left[2(10)+(41-1)(1)\right]\\\\&=(41)/(2)\left[20+(40)(1)\right]\\\\&=(41)/(2)\left[20+40\right]\\\\&=(41)/(2)\left[60\right]\\\\&=(2460)/(2)\\\\&=1230\end{aligned}`

Therefore, the sum of the counting numbers from 10 to 50 is 1230.