====== 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.

=====Question 7=====
If $a_6+a_4=6$ and $a_6-a_4=\dfrac{2}{3}$, find the arithmetic sequence. GOOD
====Solution====
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 $\dfrac{5}{3}, 2, \dfrac{7}{3}, \ldots$. GOOD
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