# If the sum of first ‘n’ odd numbers is 625 then write the value of n.

Answer

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Hint: The question hints arithmetic progression. Substitute the values in the Equation of ${{S}_{n}}$i.e, Sum of first ‘n’ number. The odd natural no. is $1,3,5\ldots \ldots $etc. Find a (first term) and d (common difference) and solve the equation to find n.

Complete step-by-step answer:

An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. We denote the common difference as ‘d’.

${{a}_{n}}$is denoted as the ${{n}^{th}}$term of an arithmetic progression.

${{S}_{n}}$is denoted as the sum of first n elements of an arithmetic series.

The sum of first n no. of an AP is given by the formula.

${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\ldots \ldots (1)$

Where a is the first term.

We have been given the sum of first ‘n’ odd number =625=${{S}_{n}}$

The odd natural no.’s are $1,3,5,7\ldots \ldots $

Here, first term, a=1

Common difference, d=2nd term-1st term=3-1=2

$\therefore $a=1, d=2, ${{S}_{n}}$=625

Putting these values in equation (1), we will get

$\begin{align}

& 625=\dfrac{n}{2}\left( 2\times 1+\left( n-1 \right)\times 2 \right) \\

& 625\times 2=n\left( 2+2n-2 \right) \\

& 625\times 2=n\left( 2n \right) \\

& \Rightarrow 625={{n}^{2}} \\

& \therefore n=\sqrt{625}=25 \\

\end{align}$

The sum of the first odd 25 terms gives the sum 625.

Note: The behaviour of AP depends upon the common difference d’. If d is positive, then the members (terms) will grow towards positive infinity or else vice versa.

Complete step-by-step answer:

An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. We denote the common difference as ‘d’.

${{a}_{n}}$is denoted as the ${{n}^{th}}$term of an arithmetic progression.

${{S}_{n}}$is denoted as the sum of first n elements of an arithmetic series.

The sum of first n no. of an AP is given by the formula.

${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\ldots \ldots (1)$

Where a is the first term.

We have been given the sum of first ‘n’ odd number =625=${{S}_{n}}$

The odd natural no.’s are $1,3,5,7\ldots \ldots $

Here, first term, a=1

Common difference, d=2nd term-1st term=3-1=2

$\therefore $a=1, d=2, ${{S}_{n}}$=625

Putting these values in equation (1), we will get

$\begin{align}

& 625=\dfrac{n}{2}\left( 2\times 1+\left( n-1 \right)\times 2 \right) \\

& 625\times 2=n\left( 2+2n-2 \right) \\

& 625\times 2=n\left( 2n \right) \\

& \Rightarrow 625={{n}^{2}} \\

& \therefore n=\sqrt{625}=25 \\

\end{align}$

The sum of the first odd 25 terms gives the sum 625.

Note: The behaviour of AP depends upon the common difference d’. If d is positive, then the members (terms) will grow towards positive infinity or else vice versa.

Last updated date: 26th May 2023

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