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.

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 $a_1=7$, $a_2=19$, $a_3=31$.

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 $a_1=1$, $a_2=5$, $a_3=9$.

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 $a_1=6$, $a_2=36$, $a_3=66$.

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