Question

1+2+3+....+100. find the AP?​

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Answer 1

Answer:

The formula for n natural no.'s =n(n+1)/2

=100(100+1)/2

=100×101/2

=50×101

=5050

Answer 2

$\large\underline{\sf{Solution-}}$

Given series is

$\rm \longmapsto\:1 + 2 + 3 + - - - + 100$

Its an AP series with

$\rm \longmapsto\:a = 1$

$\rm \longmapsto\:d = 1$

$\rm \longmapsto\: {n}^{th} \: term = 100$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Tʜᴜs,

$\rm \longmapsto\:100 = 1 + (n - 1) \times 1$

$\rm \longmapsto\:100 = 1 + n - 1$

$\bf\implies \:n = 100$

Now,

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

So, on substituting the values, we get

$\rm \longmapsto\:S_n = \dfrac{100}{2} \bigg(2 \times 1 + (100 - 1) \times 1\bigg)$

$\rm \longmapsto\:S_n =50 \times \bigg(2 + 100 - 1\bigg)$

$\rm \longmapsto\:S_n =50 \times \bigg(101\bigg)$

$\bf\implies \:S_n = 5050$

Short Cut

Sum of first n natural numbers is

$\boxed{\tt{ \displaystyle\sum_{n=1}^n\rm n \: = \: \frac{n(n \: + \: 1)}{2} \: }}$

So,

$\red{\rm \longmapsto\:\displaystyle\sum_{n=1}^{100}\rm n = \frac{100(100 + 1)}{2} = 5050}$