Question

1+3+5+7+9+11 . . . . . . . . . . . +999999

What is the sum of this AP ??​

Answer 1

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

Given AP series is

$\rm \: 1 + 3 + 5 + - - - + 999999$

having,

First term, a = 1

Common difference, d = 3 - 1 = 2

nᵗʰ term, aₙ = 999999

Let assume that number of terms in AP series be n.

We know,

↝ nᵗʰ term of an arithmetic progression is,

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

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the progression.

n is the no. of terms.

d is the common difference.

So, on substituting the values, we get

$\rm \: 999999 = 1 + (n - 1) \times 2$

$\rm \: 999999 - 1 = (n - 1) \times 2$

$\rm \: 999998 = (n - 1) \times 2$

$\rm \: n - 1 = 499999$

$\rm\implies \:n \: = \: 500000$

Now,

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic progression is,

$\begin{gathered}\:\:{\underline{{\boxed{\bf{{ \: S_n\:=\dfrac{n}{2} \bigg( \:a\:+ \: a_n \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the progression.

n is the no. of terms.

So, on substituting the values, we get

$\rm \: \: S_n\:=\dfrac{n}{2} \bigg( \:a\:+ \: a_n \bigg)$

$\rm \: \: S_n\:=\dfrac{500000}{2} \bigg( \:1\:+ \: 999999 \bigg)$

$\rm \: \: S_n\:=\dfrac{500000}{2} \bigg( \:1000000 \: \bigg)$

$\rm \: \: S_n\:=500000 \times 500000$

$\rm\implies \:\rm \: \: S_n\:=25 \times {10}^{10}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

↝ Sum of n  terms of an arithmetic progression is,

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

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the progression.

n is the no. of terms.

d is the common difference.