Question

Find the sum of the first
(i) 75 positive integers
(ii) 125 natural numbers

Answer 1

To find the sum of the first 75 positive integers, that is, $1,2,3,...,75$ , we know that this forms an AP. Hence, we will use the formula for sum of n terms of an AP, that is, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ and substitute $a=1,n=75\text{ and }d={{a}_{2}}-{{a}_{1}}=2-1=1$ in this formula for the required answer. Similarly, we can find the sum of first 125 natural numbers that is represented as $1,2,3,...,125$ using the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ . Substitute $a=1,n=125\text{ and }d={{a}_{2}}-{{a}_{1}}=2-1=1$ in the formula to get the answer.

Complete step by step answer:
We have to find the sum of the first (i) 75 positive integers and (ii)125 natural numbers.
(i) Let us find the sum of the first 75 positive integers, that is, $1,2,3,...,75$ .
We know that the above series is an AP with ${{a}_{1}}=1,{{a}_{2}}=2,n=75$ .
Hence, we can find the sum of the n terms of an AP using the formula
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]...(i)$
Where. ${{S}_{n}}$ is the sum of n terms
$n$ is the number of terms
$a$ is the first term
$d$ is the common difference.
Hence, from the series $1,2,3,...,75$ , we have
$\begin{aligned} & a=1 \\\ & n=75 \\\ \end{aligned}$
We need to find d which is the common difference of two terms. We can use the formula
$d={{a}_{2}}-{{a}_{1}}$
From the series we have ${{a}_{1}}=1\text{ and }{{a}_{2}}=2$ which are the first and second terms. Hence,
$d=2-1=1$
Now, let us substitute these in (i). That is

& {{S}_{75}}=\dfrac{75}{2}\left[ 2\times 1+\left( 75-1 \right)\times 1 \right] \\\ & \Rightarrow {{S}_{75}}=\dfrac{75}{2}\left[ 2+74 \right] \\\ \end{aligned}$$ Let us solve this. $${{S}_{75}}=\dfrac{75}{2}\times 76$$ By cancelling the common factors, we will get $${{S}_{75}}=75\times 38=2850$$ Hence, the sum of the first 75 positive integers is 2850. (ii) Let us find the sum of first 125 natural numbers, that is, $1,2,3,...,125$ . We know that the above series is an AP with $${{a}_{1}}=1,{{a}_{2}}=2,n=125$$ . Hence, we can find the sum of the AP using the formula $${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]...(i)$$ Hence, from the series $1,2,3,...,125$ , we have $\begin{aligned} & a=1 \\\ & n=125 \\\ \end{aligned}$ We need to find d which is the common difference of two terms. We can use the formula $d={{a}_{2}}-{{a}_{1}}$ From the series we have $${{a}_{1}}=1\text{ and }{{a}_{2}}=2$$ which are the first and second terms. Hence, $d=2-1=1$ Now, let us substitute these in (i). That is $$\begin{aligned} & {{S}_{125}}=\dfrac{125}{2}\left[ 2\times 1+\left( 125-1 \right)\times 1 \right] \\\ & \Rightarrow {{S}_{125}}=\dfrac{125}{2}\left[ 2+124 \right] \\\ \end{aligned}$$ Let us solve this. $${{S}_{125}}=\dfrac{125}{2}\times 126$$ By cancelling the common factors, we will get $${{S}_{125}}=125\times 63=7875$$ Hence, the sum of the first 125 natural numbers is 7875. **Note:** You may make error when writing the formula for the sum of n terms of an AP as $${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n+1 \right)d \right]$$ . We can also solve this question in an alternate way. We know that $1+2+..+n=\dfrac{n\left( n+1 \right)}{2}...(a)$ Hence, the sum of first 75 positive integers, that is $1+2+...+75$ , can be found by substituting $n=75$ in (a) . We can write this as Sum of first 75 positive integers $=\dfrac{75\left( 75+1 \right)}{2}=\dfrac{75\times 76}{2}$ Let us solve this. We will get Sum of first 75 positive integers $=75\times 38=2850$ Similarly, we can substitute $n=125$ in(a) to get the sum of the first 125 natural numbers. That is, Sum of first 125 natural numbers $=\dfrac{125\left( 125+1 \right)}{2}=\dfrac{125\times 126}{2}$ Let us solve this. We will get Sum of first 125 natural numbers $=125\times 63=7875$