Question 3 and 4, Exercise 4.2

Solutions of Question 3 and 4 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.

Find the 11th term of the arithmetic sequence $0.07,0.12,0.7, \ldots$

Solution.

Given $$0.07,0.12,0.7, \ldots$$ From this, we have $a_1 = 0.07$, $d=0.05$, $a_{11}=?$.
Now $$\begin{align*} a_n&=a_1+(n-1)d \\ \implies a_{11}&= 0.07+(11-1)(0.05)\\ &=0.07+(10)(0.05)\\ &=0.57 \end{align*}$$ Hence $a_{11}=0.57.$

The third term of an arithmetic sequence is 14 and the ninth term is -1 . Find the first four terms of the sequence.

Solution. Given: $a_3 = 14$ and $a_9 = -1$.

The nth term of the arithmetic sequence is given as $$a_n = a_1 + (n-1)d$$ Given $$\begin{align*} & a_3 = 14 \\ \implies & a_1 + 2d = 14 \quad \cdots (1) \end{align*}$$ Also $$\begin{align*} & a_9 = -1 \\ \implies & a_1 + 8d = -1 \quad \cdots (2) \end{align*}$$

Now, subtract equation (1) from equation (2): $$\begin{align*} \begin{array}{ccc} a_1 & + 8d &= -1\\ \mathop{}\limits_{-}a_1 &\mathop+\limits_{-} 2d &= \mathop{}\limits_{-}14 \\ \hline & 6d &= -15\\ \end{array}&\\ \implies \boxed{d = -\frac{5}{2}} \quad \\ \end{align*}$$

Putting the value $d$ in (1) $$\begin{align*} & a_1 + 2\left(-\frac{5}{2}\right) = 14 \\ \implies & a_1 = 14 + 5 \\ \implies & \boxed{a_1 = 19} \end{align*}$$

Now, $$\begin{align*} a_2 &= a_1 + d = 19 - \frac{5}{2} = \frac{33}{2} \\ a_3 &= a_2 + d = \frac{33}{2} - \frac{5}{2} = 14 \\ a_4 &= a_3 + d = 14 - \frac{5}{2} = \frac{23}{2} \end{align*}$$

Hence the first four terms are $19, \dfrac{33}{2}, 14, \dfrac{23}{2}$.

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