Questions & Answers

Question

Answers

[a] 6

[b] 5

[c] 4

[d] 3

[e] 2

Answer
Verified

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

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