====== Question 20, 21 and 22, Exercise 4.3 ======
Solutions of Question 20, 21 and 22 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 20=====
Find the first three terms of the arithmetic series. $a_{1}=7$, $a_{n}=139$, $S_{n}=876$.

** Solution. **

Given $a_{1}=7$, $a_{n}=139$, $S_{n}=876$. First we find $n$ and $d$.\\
As
\begin{align}
&S_n=\frac{n}{2}[a_1+a_n]\\
\implies & 876=\frac{n}{2}[7+139]\\
\implies & 1752=146n\\
\implies & n=\frac{1752}{146}=12.
\end{align}
Also we have
\begin{align}
&a_n=a_1+(n-1)d\\
\implies & 139=7+(12-1)d\\
\implies & 139-7=11d\\
\implies & d=\frac{132}{11}=12.
\end{align}
Thus
\begin{align}
&a_2=a_1+d=7+12=19\\
&a_3=a_1+2d=7+2(12)=31.\\
\end{align}
Hence $a_1=7$, $a_2=19$, $a_3=31$.







=====Question 21=====
Find the first three terms of each arithmetic series. $n=14$, $a_{n}=53$, $S_{n}=378$

** Solution. **
Given $n=14$, $a_{n}=53$, $S_{n}=378$. 
First we find $a_1$ and $d$.\\
As
\begin{align}
&S_n=\frac{n}{2}[a_1+a_n]\\
\implies & 378=\frac{14}{2}[a_1+53]\\
\implies & 378=7a_1+ 371\\
\implies & 7a_1=378-371 \\
\implies a_1=1.
\end{align}
Also we have
\begin{align}
&a_n=a_1+(n-1)d\\
\implies & 53=1+(14-1)d\\
\implies & 53-1=13d\\
\implies & d=\frac{52}{13}=4.
\end{align}
Thus
\begin{align}
&a_2=a_1+d=1+4=5\\
&a_3=a_1+2d=1+2(4)=9.\\
\end{align}
Hence $a_1=1$, $a_2=5$, $a_3=9$. GOOD





=====Question 22=====
Find the first three terms of each arithmetic series. $a_{1}=6$, $a_{n}=306$, $S_{n}=1716$.

** Solution. **

Given $a_{1}=6$, $a_{n}=306$, $S_{n}=1716$. 
First we find $n$ and $d$.\\
As
\begin{align}
&S_n=\frac{n}{2}[a_1+a_n]\\
\implies & 1716=\frac{n}{2}[6+306]\\
\implies & 3432=312n\\
\implies & n=\frac{3432}{312}=11.
\end{align}
Also we have
\begin{align}
&a_n=a_1+(n-1)d\\
\implies & 306=6+(11-1)d\\
\implies & 306-6=10d\\
\implies & d=\frac{300}{10}=30.
\end{align}
Thus
\begin{align}
&a_2=a_1+d=6+30=36\\
&a_3=a_1+2d=6+2(30)=66.\\
\end{align}
Hence $a_1=6$, $a_2=36$, $a_3=66$. GOOD






====Go to ====

<text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-3-p9|< Question 17, 18 & 19]]</btn></text>
<text align="right"><btn type="success">[[math-11-nbf:sol:unit04:ex4-3-p11|Question 23 & 24 >]]</btn></text>