====== 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 A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 3=====
Find the numbers of terms in arithmetic progression $6,9,12, \ldots, 78$.
====Solution====
Here $a_1=6$ and $d=9-6=3$ and $a_n=78$.\\
We know that $$a_n=a_1+(n-1) d$$
This gives
\begin{align}&78=6+(n-1) 3 \\
\implies &3(n-1)=78-6 \\
\implies &n-1=\dfrac{72}{3} \\
\implies &n=24+1=25.\end{align}
Thus, the number of terms in given progression are $25$. GOOD

=====Question 4=====
The $n$th term of sequence is given by $a_n=2n+7$. Show that it is an arithmetic progression. Also find its 7th term.
====Solution====
Given that $$a_n=2 n+7. --- (1)$$
Then
\begin{align}a_{n+1}=2(n+1)+7=2 n+9.\end{align}
Take
\begin{align}
d&=a_{n+1}-a_n\\
&=2n+9-2n-7=2\end{align}
This gives every two terms of the sequence has same distance $2$, hence it is an A.P. 

Putting $n=7$ in (1), we get
$$a_7=2(7)+7=14+7=21.$$
Hence the 7th term of given AP is 21. GOOD
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