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.

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

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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