# If the ${{n}^{th}}$ term of an A.P. is $(2n+1)$, find the sum of the first $n$ terms of the A.P.

Last updated date: 27th Mar 2023

•

Total views: 206.1k

•

Views today: 3.83k

Answer

Verified

206.1k+ views

Hint: The given question is related to arithmetic progression. Try to recall the formulae related to ${{n}^{th}}$ term of an arithmetic progression and the sum of first $n$ terms of an arithmetic progression.

Before proceeding with the solution, we must know the concept of arithmetic progression. Arithmetic progression is a series of numbers in which the difference between any two consecutive numbers is always constant.

Let’s consider an arithmetic progression ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}.....$ .

We know, the difference between consecutive terms is constant. Let the difference between consecutive numbers be $d$.

So, $d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}...$

So , we can write ${{a}_{2}}={{a}_{1}}+d$ and ${{a}_{3}}={{a}_{2}}+d={{a}_{1}}+2d$ .

Similarly, ${{a}_{4}}={{a}_{3}}+d={{a}_{1}}+3d$ .

Following the pattern, we can say that the ${{n}^{th}}$ term of the arithmetic progression is given as ${{a}_{n}}={{a}_{1}}+(n-1)d$ .

Now, let the sum of the first $n$ terms of an arithmetic progression be denoted as ${{S}_{n}}$.

So, ${{S}_{n}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}$.

Substituting the values of ${{a}_{1}},{{a}_{2}},{{a}_{3}}....{{a}_{n}}$ in the expression of ${{S}_{n}}$ , we get

${{S}_{n}}={{a}_{1}}+\left( {{a}_{1}}+d \right)+\left( {{a}_{1}}+2d \right)+\left( {{a}_{1}}+3d \right)+.....+\left( {{a}_{1}}+(n-1)d \right)$

$\Rightarrow {{S}_{n}}=n{{a}_{1}}+d(1+2+3+...+(n-1))$ .

Now, we know, the sum of first $n$ natural numbers is given as $\dfrac{n(n+1)}{2}$ .

So, the sum of first $(n-1)$ natural numbers is $\dfrac{(n-1)n}{2}$

So, ${{S}_{n}}=n{{a}_{1}}+\dfrac{n(n-1)}{2}d$

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2{{a}_{1}}+(n-1)d \right]$ .

Now, we are given an arithmetic progression whose ${{n}^{th}}$ term is given as $(2n+1)$.

To find the first term of the progression, we will substitute $n=1$ in the expression of the ${{n}^{th}}$ term.

So, the first term is given as ${{a}_{1}}=(2\times 1)+1=3$.

Now, to find the common difference, we must know the value of ${{2}^{nd}}$ term of the progression. To find the value of the second term, we will substitute $n=2$ in the expression of the ${{n}^{th}}$ term.

So, the second term is given as ${{a}_{2}}=\left( 2\times 2 \right)+1=5$.

Now, the common difference $d={{a}_{2}}-{{a}_{1}}=5-3=2$.

So, we have an arithmetic progression with the first term ${{a}_{1}}=3$ and common difference $d=2$ .

So, the sum of the first $n$ terms of the given arithmetic progression is given as ${{S}_{n}}=\dfrac{n}{2}\left[ (2\times 3)+(n-1)2 \right]$

$=\dfrac{n}{2}\left( 6+2n-2 \right)$

$=\dfrac{n}{2}(4+2n)$

$=n(n+2)$

Hence, the sum of the first $n$ terms of an arithmetic progression, whose ${{n}^{th}}$ term is given by $(2n+1)$, is given as ${{S}_{n}}=n(n+2)$.

Note: The common difference of an arithmetic progression is given as $d={{a}_{2}}-{{a}_{1}}$ and not $d={{a}_{1}}-{{a}_{2}}$. Students generally get confused and make this mistake.

Before proceeding with the solution, we must know the concept of arithmetic progression. Arithmetic progression is a series of numbers in which the difference between any two consecutive numbers is always constant.

Let’s consider an arithmetic progression ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}.....$ .

We know, the difference between consecutive terms is constant. Let the difference between consecutive numbers be $d$.

So, $d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}...$

So , we can write ${{a}_{2}}={{a}_{1}}+d$ and ${{a}_{3}}={{a}_{2}}+d={{a}_{1}}+2d$ .

Similarly, ${{a}_{4}}={{a}_{3}}+d={{a}_{1}}+3d$ .

Following the pattern, we can say that the ${{n}^{th}}$ term of the arithmetic progression is given as ${{a}_{n}}={{a}_{1}}+(n-1)d$ .

Now, let the sum of the first $n$ terms of an arithmetic progression be denoted as ${{S}_{n}}$.

So, ${{S}_{n}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}$.

Substituting the values of ${{a}_{1}},{{a}_{2}},{{a}_{3}}....{{a}_{n}}$ in the expression of ${{S}_{n}}$ , we get

${{S}_{n}}={{a}_{1}}+\left( {{a}_{1}}+d \right)+\left( {{a}_{1}}+2d \right)+\left( {{a}_{1}}+3d \right)+.....+\left( {{a}_{1}}+(n-1)d \right)$

$\Rightarrow {{S}_{n}}=n{{a}_{1}}+d(1+2+3+...+(n-1))$ .

Now, we know, the sum of first $n$ natural numbers is given as $\dfrac{n(n+1)}{2}$ .

So, the sum of first $(n-1)$ natural numbers is $\dfrac{(n-1)n}{2}$

So, ${{S}_{n}}=n{{a}_{1}}+\dfrac{n(n-1)}{2}d$

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2{{a}_{1}}+(n-1)d \right]$ .

Now, we are given an arithmetic progression whose ${{n}^{th}}$ term is given as $(2n+1)$.

To find the first term of the progression, we will substitute $n=1$ in the expression of the ${{n}^{th}}$ term.

So, the first term is given as ${{a}_{1}}=(2\times 1)+1=3$.

Now, to find the common difference, we must know the value of ${{2}^{nd}}$ term of the progression. To find the value of the second term, we will substitute $n=2$ in the expression of the ${{n}^{th}}$ term.

So, the second term is given as ${{a}_{2}}=\left( 2\times 2 \right)+1=5$.

Now, the common difference $d={{a}_{2}}-{{a}_{1}}=5-3=2$.

So, we have an arithmetic progression with the first term ${{a}_{1}}=3$ and common difference $d=2$ .

So, the sum of the first $n$ terms of the given arithmetic progression is given as ${{S}_{n}}=\dfrac{n}{2}\left[ (2\times 3)+(n-1)2 \right]$

$=\dfrac{n}{2}\left( 6+2n-2 \right)$

$=\dfrac{n}{2}(4+2n)$

$=n(n+2)$

Hence, the sum of the first $n$ terms of an arithmetic progression, whose ${{n}^{th}}$ term is given by $(2n+1)$, is given as ${{S}_{n}}=n(n+2)$.

Note: The common difference of an arithmetic progression is given as $d={{a}_{2}}-{{a}_{1}}$ and not $d={{a}_{1}}-{{a}_{2}}$. Students generally get confused and make this mistake.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE