# 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.

## Question 1

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 $a_{15}=46$.

## Question 2

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 $a_{15} = 44$.

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