====== Question 3 Exercise 5.3 ======
Solutions of Question 3 of Exercise 5.3 of Unit 05: Miscullaneous 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 $n$ term and sum to $n$ terms each of the series $4+10+18+28+40+\ldots$
====Solution====
We use the method of difference as:
\begin{align}
& a_2-a_1=10-4=6 \\
& a_3-a_2=18-10=8 \\
& a_4-a_3=28-18=10 \\
& \text {... ... ... } \\
& \text {... ... ... } \\
& a_n-a_{n \quad 1}=(\mathrm{n}-1) \text { term of the sequence } \end{align}
$6,10,8, \ldots$ which is a A.P. Adding column wise, we get
\begin{align}& a_n-a_1=6+10+8-\ldots +(n-1) \text { terms } \\
& =\dfrac{n-1}{2}[2 \cdot 6+(n-2) \cdot 2] \\
& =\dfrac{n-1}{2}[12+2 n-4] \\
& =\dfrac{n-1}{2}[2 n+8]=2 \cdot \dfrac{n-1}{2} \cdot[n+4] \\
& \Rightarrow a_n=n^2+3 n-4+a_1 \\
& \Rightarrow a_n=n^2+3 n-4+4 \quad \because a_1=4 \\
& \Rightarrow a_n=n^2+3 n\end{align}
Taking summation of the both sides
\begin{align}\sum_{r=1}^n a_r&=\sum_{r=1}^n r^2+3 \sum_{r=1}^n r \\
& =\dfrac{n(n+1)(2 n+1)}{6}+3 \dfrac{n(n+1)}{2} \\
& =\dfrac{n(n+1)}{2}[\dfrac{2 n+1}{3}+3] \\
& =\dfrac{n(n+1)}{2}[\dfrac{2 n+1+9}{3}] \\
& =\dfrac{n(n+1)}{2}[\dfrac{2 n+10}{3}] \\
& =\dfrac{n}{3}(n+1)(n+5)\\
\Rightarrow \sum_{r=1}^n a_r&=\dfrac{n}{3}(n+1)(n+5)\\
\text{Hence}\quad n^2+3 n;\dfrac{n}{3}(n+1)(n+5)\end{align}




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