Question 1, Exercise 4.2

Solutions of Question 1 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 first four terms of the arithmetic sequence with $a_{1}=4, d=3$

Solution.

Given: $a_1= 4$, $d=3$.
The general term of an arithmetic sequence is: $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&=4+(2-1)3=4+3=7\\ a_3 &= 4+ (3-1) 3 = 4 + 6 = 10\\ a_4&=4+(4-1)3=4+9=13 \end{align*} Hence $a_1=4$, $a_2=7$, $a_3=10$, $a_4=13$.

Find the first four terms of the arithmetic sequence with $a_1=7$, $d=5$

Solution.

Given: $a_1= 7$, $d=5$.
The general term of an arithmetic sequence is: $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&=7+(2-1)(5)=7+5=12\\ a_3 &= 7+ (3-1)(5) = 7 + 10 = 17\\ a_4&=7+(4-1)(5)=7+15=22 \end{align*} Hence $a_1=7$, $a_2=12$, $a_3=17$, $a_4=22$.

Find the first four terms of each arithmetic sequence. $a_{1}=16$, $d=-2$.

Solution.

Given: $a_1= 16$, $d=-2$.
We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&=16+(2-1)(-2)=16-2=14\\ a_3 &= 16+ (3-1)(-2) = 16 - 4 = 12\\ a_4&=16+(4-1)(-2)=16-6=10 \end{align*} Hence $a_1=16$, $a_2=14$, $a_3=12$, $a_4=10$.

Find the first four terms of the arithmetic sequence. $a_1=38$, $d=-4$.

Solution.

Given: $a_1= 38$, $d=-4$.
We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&=38+(2-1)(-4)=38-4=34\\ a_3 &= 38+ (3-1)(-4) = 38 - 8 = 30\\ a_4&=38+(4-1)(-4)=38-12=26 \end{align*} Hence $a_1=38$, $a_2=34$, $a_3=30$, $a_4=26$.

Find the first four terms of each arithmetic sequence. $a_{1}=\frac{3}{4}, d=\frac{1}{4}$

Solution.

Given: $a_1=\frac{3}{4}$, $d=\frac{1}{4}$.
We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&=\frac{3}{4}+(2-1)\cdot\frac{1}{4}=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1\\ a_3 &= \frac{3}{4}+(3-1)\cdot\frac{1}{4}=\frac{3}{4}+\frac{2}{4}=\frac{5}{4}\\ a_4&=\frac{3}{4}+(4-1)\cdot\frac{1}{4}=\frac{3}{4}+\frac{3}{4}=\frac{6}{4}=1.5 \end{align*} Hence $a_1=\frac{3}{4}$, $a_2=1$, $a_3=\frac{5}{4}$, $a_4=\frac{6}{4}=1.5$.

Find the first four terms of each arithmetic sequence. $a_{1}=\frac{3}{8}, d=\frac{5}{8}$

Solution.

Given: $a_1=\frac{3}{8}$, $d=\frac{5}{8}$.
We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2 &= \frac{3}{8} + (2-1) \cdot \frac{5}{8} = \frac{3}{8} + \frac{5}{8} = \frac{8}{8} = 1\\ a_3 &= \frac{3}{8} + (3-1) \cdot \frac{5}{8} = \frac{3}{8} + 2 \cdot \frac{5}{8} = \frac{3}{8} + \frac{10}{8} = \frac{13}{8}\\ a_4 &= \frac{3}{8} + (4-1) \cdot \frac{5}{8} = \frac{3}{8} + 3 \cdot \frac{5}{8} = \frac{3}{8} + \frac{15}{8} = \frac{18}{8} = \frac{9}{4} \end{align*} Hence $a_1=\frac{3}{8}$, $a_2=1$, $a_3=\frac{13}{8}$, $a_4=\frac{9}{4}$.

Question 2 >