====== Question 1 Exercise 4.3 ======
Solutions of Question 1 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 1(i)=====
Find indicated term and sum of the indicated number of terms in arithmetic sequence: $9,7,5,3, \ldots$; 20th term; 20 terms. GOOD
====Solution====
Let $a_1$ be first term and $d$ be common difference of given A.P. Then
\begin{align}&a_1=9 \\ 
&d=7-9=-2 \\
&n=20.
\end{align}
We know that
\begin{align}&a_n=a_1+(n-1)d \\
\implies &a_20=9+(20-1)(-2)=-29.
\end{align}
Assume $S_n$ represents the sum of first $n$ terms of A.P. Then
\begin{align}
S_n&=\dfrac{n}{2}[a_1+a_n], \\
\implies S_{20}&=\dfrac{20}{2}[9-29] \\
&=10(-20)=-200 \end{align}
Hence 20th term is -29 and sum of first 20 terms is -200. GOOD

=====Question 1(ii)=====
Find indicated term and sum of the indicated number of terms in case of arithmetic sequence: $3, \dfrac{8}{3}, \dfrac{7}{3}, 2, \ldots$; 11th term; 11 terms. GOOD
====Solution====
Let $a_1$ be first term and $d$ be common difference of given A.P. Then
\begin{align}&a_1=3 \\ 
&d=\dfrac{8}{3}-3=-\dfrac{1}{3} \\
&n=11.
\end{align}
We know that
$$a_n=a_1+(n-1) d,$$
This gives
$$a_{11}=3+10\left(-\dfrac{1}{3}\right)=-\dfrac{1}{3}$$
Assume $S_n$ represents the sum of first $n$ terms of A.P. Then
\begin{align}
S_n&=\dfrac{n}{2}[a_1+a_n], \\
\implies S_{11}&=\dfrac{11}{2}\left[3-\dfrac{1}{3}\right] \\
&=\dfrac{11}{2} \cdot \dfrac{9-1}{3} \\&=\dfrac{44}{3} \end{align}
Hence 11th term is $-\dfrac{1}{3}$ and sum of first 11 terms is $\dfrac{44}{3}$. GOOD

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