# Question 1 and 2, Exercise 4.3

Solutions of Question 1 and 2 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 1

Find the sum of series: $4+7+10+13+16+19+22+25$

** Solution. **
Given: $4+7+10+13+16+19+22+25$.

As the given series is arithmetic series with $a_1=4$, $d=7-4=3$, $n=8$, so \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d]\\ \implies S_8&=\frac{8}{2}[2(4)+(8-1)(3)]\\ &=4[8+7\times 3] = 116 \end{align} Hence sum of given series is 116.

## Question 2

Find the sum of series. $a_{1}=2$, $a_{n}=200$, $n=100$

Statement is logically incorrenct.

Find the sum of arithematic series with: $a_{1}=2$, $a_{n}=200$, $n=100$

** Solution. **

Given $a_{1}=2$, $a_{n}=200$, $n=100$.

Let $S_n$ represents sum of arithematic series. Then
\begin{align}
S_n&=\frac{n}{2}[a_1+a_n] \\
\implies S_{100}&=\frac{100}{2}[2+200]\\
&=10100.
\end{align}
Hence $S_{100}=10100$.

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