Question 7 Exercise 4.2

Solutions of Question 7 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.

If $a_6+a_4=6$ and $a_6-a_4=\dfrac{2}{3}$, find the arithmetic sequence.

Let $a_1$ be first term and $d$ be common difference of A.P. As given $$\begin{align} &a_6+a_4=6 \\ \implies & a_1+5d+a_1+3d=6\\ \implies & 2a_1+8d=6\\ \implies & a_1+4d=3 --- (1) \end{align}$$ Also, we have given $$\begin{align} &a_6-a_4=\dfrac{2}{3} \\ \implies & a_1+5d-a_1-3d=\dfrac{2}{3}\\ \implies & 2d=\dfrac{2}{3}\\ \implies & d=\dfrac{1}{3} \end{align}$$ Using the value of $d$ in (1), we get $$\begin{align} &a_1+4\left(\dfrac{1}{3}\right)=3\\ \implies &a_1=3-\dfrac{4}{3}=\dfrac{5}{3}. \end{align}$$ Also $$\begin{align} a_2&=a_1+d\\ &=\dfrac{5}{3}+\dfrac{1}{3}=2\end{align}$$ and $$\begin{align} a_3&=a_1+2d\\ &=\dfrac{5}{3}+2\dfrac{1}{3}=\dfrac{7}{3}\end{align}$$ Hence the required A.P is $\dfrac{5}{3}, 2, \dfrac{7}{3}, \ldots$.

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