# Question 2, Exercise 4.2

Solutions of Question 2 of Exercise 4.2 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 2(i)

Find the next three terms of each arithmetic sequence. $5,9,13, \ldots$

** Solution. **

Give: $$5, 9, 13, \ldots $$
Thus $a_1=5$, $d=9-5=4$.

Now
$$a_n=a_1+(n-1)d.$$
So, we have
\begin{align*}
a_4 &=5+(4-1)(4)=5+12=17\\
a_5 &=5+(5-1)(4)=5+16=21\\
a_6 &=5+(6-1)(4)=5+20=25
\end{align*}
Thus, the next three terms of the sequence are $17$, $21$, $25$.

## Question 2(ii)

Find the next three terms of each arithmetic sequence. $11,14,17, \ldots$

** Solution. **

Given: $$11, 14, 17, \ldots$$
Thus $a_1=11$, $d=14-11=3$.

Now
$$a_n=a_1+(n-1)d.$$
So, we have
\begin{align*}
a_4 &= 11 + (4-1) \cdot 3 = 11 + 9 = 20\\
a_5 &= 11 + (5-1) \cdot 3 = 11 + 12 = 23\\
a_6 &= 11 + (6-1) \cdot 3 = 11 + 15 = 26
\end{align*}
Thus, the next three terms of the sequence are $20$, $23$, $26$.

## Question 2(iii)

Find the next three terms of each arithmetic sequence. $\dfrac{1}{2}, \dfrac{3}{2}, \dfrac{5}{2}, \ldots$

** Solution. **

The given sequence is $$\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \ldots$$ First, we find the common difference: \begin{align*} d &= \frac{3}{2} - \frac{1}{2} = 1 \end{align*} The arithmetic sequence is: $$a_n = a_{n-1} + d$$ Now, let's calculate the next three terms: \begin{align*} a_3 &= \frac{5}{2}\\ a_4 &= a_3 + d = \frac{5}{2} + 1 = \frac{7}{2}\\ a_5 &= a_4 + d = \frac{7}{2} + 1 = \frac{9}{2}\\ a_6 &= a_5 + d = \frac{9}{2} + 1 = \frac{11}{2} \end{align*} Thus, the next three terms of the sequence are $ \dfrac{7}{2}, \dfrac{9}{2}, \dfrac{11}{2} $.

## Question 2(iv)

Find the next three terms of the arithmetic sequence. $-5.4,-1.4,-2.6, \ldots$

** Solution. **

Given: $$-5.4, -1.4, 2.6, \ldots$$ Thus, \(a_1 = -5.4\), and \(d = -1.4 - (-5.4) = 4\). Now $$a_n = a_1 + (n - 1)d.$$ So, we have \[ \begin{aligned} a_4 &= -5.4 + (4 - 1)(4) = -5.4 + 12 = 6.6, \\ a_5 &= -5.4 + (5 - 1)(4) = -5.4 + 16 = 10.6, \\ a_6 &= -5.4 + (6 - 1)(4) = -5.4 + 20 = 14.6. \end{aligned} \] Thus, the next three terms of the sequence are \(6.6\), \(10.6\), and \(14.6\).

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