====== Question 2 Exercise 4.3 ======
Solutions of Question 2 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 2(i)=====
Some of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that is missing: $a_1=2, n=17, d=3$. GOOD
====Solution====
Given: $a_1=2, n=17, d=3$ \\
We need to find $a_{17}$ and $S_{17}$. As we know
$$a_{n}=a_1+(n-1)d.$$
Thus
$$a_{17}=2+(17-1)(3)=50.$$
Also
$$S_n=\dfrac{n}{2}[a_1+a_n]$$
Thus
\begin{align}S_{17}&=\dfrac{17}{2}(a_1+a_17) \\
&=\dfrac{17}{2}(2+50)=442.\end{align}
Hence $a_{17}=50$ and $S_{17}=442$. GOOD

=====Question 2(ii)=====
Some of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that are missing $a_1=-40, S_{21}=210$. GOOD
====Solution====
Given: $a_1=-40$ and $S_{21}=210$.\\
So we have $n=21$ and we have to find $a_{21}$ and $d$. As
\begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\
\implies &210=\dfrac{21}{2}(-40+a_{21}) \\
\implies &-40+a_{21}=\dfrac{2 \times 210}{21}=20 \\
\implies &a_{21}=20+40=60.\end{align}
Also $a_{21}=a_1+20 d$, then\\
\begin{align}&20d=60-(-40)=100 \\
\implies &d=\dfrac{100}{20}=5
\end{align}
Hence $a_{21}=60$, $d=5$ and $n=21$. GOOD
=====Question 2(iii)=====
Some of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that are missing $a_1=-7, d=8, S_n=225$. GOOD
====Solution====
Given: $a_1=-7, d=8, S_n=225$, we have to find $n$ and $a_n$.
We know that
$$S_n=\dfrac{n}{2}[2 a_1+(n-1) d].$$
Thus, we have
\begin{align}
& 225=\dfrac{n}{2}[2 \cdot(-7)+(n-1) \cdot 8], \\
\implies & n[-14+8(n-1)]=2 \times 225 \\
\implies & -14 n+8 n(n-1)=450 \\
\implies & 8 n^2-8 n-14 n=450 \\
\implies & 8 n^2-22 n-450=0 \\
\implies & 4 n^2-11 n-225=0\end{align}
This is quadratic equation with $a=4, b=-11$ and $c=-225$, then
\begin{align}n&=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
n&=\dfrac{11 \pm \sqrt{(-11)^2-4(4)(-225)}}{2.4} \\
\implies n&=\dfrac{11 \pm \sqrt{121+3600}}{8} \\
\implies n&=\dfrac{11 \pm \sqrt{3721}}{8}\\
&=\dfrac{11 \pm 61}{8} \\
\implies n&=\dfrac{11+61}{8} \text { or } n=\dfrac{11-61}{8} \\
\implies n&=9 \text { or } n=-\dfrac{50}{8}\end{align}
Since $n$ cannot be negative or in fraction, thus $n=9$.
Now $$a_9=a_1+8 d=-7+8(8)=57.$$
Hence $n=9$ and $a_{9}=57$. GOOD

=====Question 2(iv)=====
Some of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that are missing: $a_n=4, S_{15}=30$. GOOD
====Solution====
Given: $a_n=4, S_{15}=30$.\\
Thus we have $n=15$ and we have to find $a_1$ and $d$.
We know that $$S_n=\dfrac{n}{2}[a_1+a_n],$$
then we have
\begin{align}&S_{15}=\dfrac{15}{2}[a_1+a_{15}]\\
\implies &\dfrac{15}{2}[a_1+4]=30 \\
\implies &a_1+4=\dfrac{30 \times 2}{15}=4 \\
\implies &a_1=4-4=0. \end{align}
Also
\begin{align}&a_{15}=a_1+14d \\
\implies &4=0+14d \\
\implies &d=\dfrac{4}{14}=\dfrac{2}{7}.\end{align}
Hence $a_1=0$, $n=15$ and $d=\dfrac{2}{7}$. GOOD


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