Question 15 and 16, Exercise 4.3
Solutions of Question 15 and 16 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 15
Find $S_n$ for arithmetic series. $a_{1}=91$, $d=-4$, $a_{n}=15$.
Solution.
Given: $a_{1}=91$, $d=-4$, $a_{n}=15$, $n=?$.
We have
\begin{align}
& a_n=a_1+(n-1)d \\
\implies & 15=91+(n-1)(-4) \\
\implies & 15=91-4n+4 \\
\implies & 4n=95-15 \\
\implies & 4n = 80\\ \implies & n = 20.
\end{align}
Now
\begin{align}
S_n&=\frac{n}{2}[a_1+a_n] \\
\implies S_{20}&=\frac{20}{2}[91+15]\\
&=10\times 106\\
&=1060.
\end{align}
Hence $S_{20}=1060$. 
Question 16
Find $S_n$ for arithmetic series. $d=-4$, $n=9$, $a_{n}=27$.
Solution.
Given: $d=-4$, $n=9$, $a_{n}=27$, $a_1=?$.
We have
\begin{align}
& a_n=a_1+(n-1)d \\
\implies & 27=a_1+(9-1)(-4) \\
\implies & 27=a_1-32 \\
\implies & a_1=27+32 =59.
\end{align}
Now
\begin{align}
S_n&=\frac{n}{2}[a_1+a_n] \\
\implies S_{9}&=\frac{9}{2}[59+27]\\
&=\frac{9}{2}\times 86\\
&=387.
\end{align}
Hence $S_9=387$.
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