Question 1 and 2, Exercise 4.1

Solutions of Question 1 and 2 of Exercise 4.1 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.

The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$. $$a_{n}=3 n+1$$

Solution. Given $$a_{n}=3 n+1$$ Then $$\begin{align*} a_1 &= 3(1) + 1 = 3 + 1 = 4\\ a_2 &= 3(2) + 1 = 6 + 1 = 7\\ a_3 &= 3(3) + 1 = 9 + 1 = 10\\ a_4 &= 3(4) + 1 = 12 + 1 = 13\\ \end{align*}$$ Now $$\begin{align*} a_{10} &= 3(10) + 1 = 30 + 1 = 31\\ a_{15} &= 3(15) + 1 = 45 + 1 = 46. \end{align*}$$ Hence $a_1=4$, $a_2=7$, $a_3=10$, $a_4=13$, $a_{10}=31$ and $a_{15}=46$.

The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term, $a_{15}$. $$a_{n}=3 n-1$$

Solution. Given: $$a_{n}=3 n-1.$$ Then $$\begin{align*} a_1 &= 3(1) - 1 = 3 - 1 = 2\\ a_2 &= 3(2) - 1 = 6 - 1 = 5\\ a_3 &= 3(3) - 1 = 9 - 1 = 8\\ a_4 &= 3(4) - 1 = 12 - 1 = 11\\ \end{align*}$$ Now $$\begin{align*} a_{10} &= 3(10) - 1 = 30 - 1 = 29\\ a_{15} &= 3(15) - 1 = 45 - 1 = 44 \end{align*}$$ So, the first four terms are $a_1=2$, $a_2=5$, $a_3=8$, $a_4=11$, $a_{10} = 29$ and $a_{15} = 44$.

Question 3 & 4 >