Answer

Verified

481.2k+ views

Hint: Use the formula for the nth term of A.P, i.e. ${{a}_{n}}=a+(n-1)d$. Form two equations in terms of a and d and solve the system of equations. The solution of the system will give the value of a.

Complete step by step solution:

An arithmetic progression is a sequence of numbers in which adjacent numbers differ by a constant factor.

It is represented in the form $a,a+d,a+2d,\ldots $

a is known as the first term of the A.P and d the common difference.

Let “a” be the first term and “d” the common difference of the A.P. Then we have

${{a}_{21}}+{{a}_{22}}+{{a}_{23}}=261\text{ (i)}$ { Because the sum of the last three terms is 261.}

The three middle terms are ${{a}_{11}},{{a}_{12}}$ and \[{{a}_{13}}\] . So, we have

${{a}_{11}}+{{a}_{12}}+{{a}_{13}}=141\text{ (ii)}$

From equation (i) we have $a+\left( 21-1 \right)d+a+\left( 22-1 \right)d+a+\left( 23-1 \right)d=261$

$\begin{align}

& \Rightarrow a+20d+a+21d+a+22d=261 \\

& \Rightarrow 3a+63d=261\text{ (iii)} \\

\end{align}$

From equation (ii) we have $a+\left( 11-1 \right)d+a+\left( 12-1 \right)d+a+\left( 13-1 \right)d=141$

$\begin{align}

& \Rightarrow a+10d+a+11d+a+12d=141 \\

& \Rightarrow 3a+33d=141\text{ (iv)} \\

\end{align}$

Subtracting equation (iv) from equation (iii), we get

$\begin{align}

& 3a+63d-\left( 3a+33d \right)=261-141 \\

& \Rightarrow 3a+63d-3a-33d=120 \\

& \Rightarrow 30d=120 \\

\end{align}$

Dividing both sides by 30, we get

$\dfrac{30d}{30}=\dfrac{120}{30}=4$

i.e. d = 4.

Substituting the value of d in equation (iv), we get

$\begin{align}

& 3a+33\times 4=141 \\

& \Rightarrow 3a+132=141 \\

& \Rightarrow 3a=9 \\

& \Rightarrow a=3 \\

\end{align}$

Hence the first term of the A.P = 3.

Note: [a] When asked to calculate m middle terms of a sequence of n numbers we first find the middle term. We know that if n is odd, then $\left( \dfrac{n+1}{2} \right)th$ terms is the middle term. If n is even, then $\left( \dfrac{n}{2} \right)th$ and $\left( \dfrac{n}{2}+1 \right)th$ terms are the middle terms. We then move an equal number of steps to the left and the right of the middle term to get the other middle terms.

Consider the case in which we have 23 terms in a sequence, and we have to find the three middle terms. Middle term of the sequence will be $\dfrac{23+1}{2}th=\dfrac{24}{2}th=12th$ term. Moving one step to the left and one to the right of the middle term, we have the three middle terms are 11th, 12th and 13th, which is the same as done in the above solution.

[b] Some important formulae in A.P are:

[1] ${{a}_{n}}=a+\left( n-1 \right)d$

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

[3] ${{S}_{n}}=\dfrac{n}{2}\left[ 2l-\left( n-1 \right)d \right]$ where l is the last term.

[4] ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$

[5] ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$

Complete step by step solution:

An arithmetic progression is a sequence of numbers in which adjacent numbers differ by a constant factor.

It is represented in the form $a,a+d,a+2d,\ldots $

a is known as the first term of the A.P and d the common difference.

Let “a” be the first term and “d” the common difference of the A.P. Then we have

${{a}_{21}}+{{a}_{22}}+{{a}_{23}}=261\text{ (i)}$ { Because the sum of the last three terms is 261.}

The three middle terms are ${{a}_{11}},{{a}_{12}}$ and \[{{a}_{13}}\] . So, we have

${{a}_{11}}+{{a}_{12}}+{{a}_{13}}=141\text{ (ii)}$

From equation (i) we have $a+\left( 21-1 \right)d+a+\left( 22-1 \right)d+a+\left( 23-1 \right)d=261$

$\begin{align}

& \Rightarrow a+20d+a+21d+a+22d=261 \\

& \Rightarrow 3a+63d=261\text{ (iii)} \\

\end{align}$

From equation (ii) we have $a+\left( 11-1 \right)d+a+\left( 12-1 \right)d+a+\left( 13-1 \right)d=141$

$\begin{align}

& \Rightarrow a+10d+a+11d+a+12d=141 \\

& \Rightarrow 3a+33d=141\text{ (iv)} \\

\end{align}$

Subtracting equation (iv) from equation (iii), we get

$\begin{align}

& 3a+63d-\left( 3a+33d \right)=261-141 \\

& \Rightarrow 3a+63d-3a-33d=120 \\

& \Rightarrow 30d=120 \\

\end{align}$

Dividing both sides by 30, we get

$\dfrac{30d}{30}=\dfrac{120}{30}=4$

i.e. d = 4.

Substituting the value of d in equation (iv), we get

$\begin{align}

& 3a+33\times 4=141 \\

& \Rightarrow 3a+132=141 \\

& \Rightarrow 3a=9 \\

& \Rightarrow a=3 \\

\end{align}$

Hence the first term of the A.P = 3.

Note: [a] When asked to calculate m middle terms of a sequence of n numbers we first find the middle term. We know that if n is odd, then $\left( \dfrac{n+1}{2} \right)th$ terms is the middle term. If n is even, then $\left( \dfrac{n}{2} \right)th$ and $\left( \dfrac{n}{2}+1 \right)th$ terms are the middle terms. We then move an equal number of steps to the left and the right of the middle term to get the other middle terms.

Consider the case in which we have 23 terms in a sequence, and we have to find the three middle terms. Middle term of the sequence will be $\dfrac{23+1}{2}th=\dfrac{24}{2}th=12th$ term. Moving one step to the left and one to the right of the middle term, we have the three middle terms are 11th, 12th and 13th, which is the same as done in the above solution.

[b] Some important formulae in A.P are:

[1] ${{a}_{n}}=a+\left( n-1 \right)d$

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

[3] ${{S}_{n}}=\dfrac{n}{2}\left[ 2l-\left( n-1 \right)d \right]$ where l is the last term.

[4] ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$

[5] ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

The states of India which do not have an International class 10 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you graph the function fx 4x class 9 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

Name the three parallel ranges of the Himalayas Describe class 9 social science CBSE