====== Question 7 and 8, Exercise 4.2 ======
Solutions of Question 7 and 8 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. 


=====Question 7=====
Which term of the sequence $-6,-2,2, \ldots$ is $70$?

** Solution. **

Given $-6,-2,2, \ldots$ is an arithmetic sequence.

Here $a_1=-6$, $d=-2+6=4$, $a_n=70$, $n=?$.

The nth term of the arithmetic sequence is given as
$$a_n=a_1+(n-1)d.$$
This gives
\begin{align*}
&70=-6+(n-1)4\\
\implies &70=-6+4n-4\\
\implies &70=4n-10\\
\implies &4n=80\\
\implies & n=20
\end{align*}
Hence $a_{20}=70$. GOOD

=====Question 8=====
Which term of the sequence $\dfrac{5}{2}, \dfrac{3}{2}, \dfrac{1}{2}, \ldots$ is $-\dfrac{105}{2}$?

** Solution. **

Given: $\dfrac{5}{2}, \dfrac{3}{2}, \dfrac{1}{2}, \ldots$ is an arithmetic sequence and $a_n = -\dfrac{105}{2}$.

Here $a_1 = \dfrac{5}{2}$, $d = \dfrac{3}{2} - \dfrac{5}{2} = -1$, $n=?$.

The nth term of the arithmetic sequence is given as
$$a_n = a_1 + (n-1)d.$$
This gives
\begin{align*}
& -\frac{105}{2} = \frac{5}{2} + (n-1)(-1)\\
\implies & -\frac{105}{2} = \frac{5}{2} - n + 1\\
\implies & -\frac{105}{2} = \frac{7}{2} - n\\
\implies & n = \frac{7}{2} + \frac{105}{2}\\
\implies & n = 56
\end{align*}

Hence $a_{56} = -\dfrac{105}{2}$. GOOD




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