Question 6 Exercise 5.3

Solutions of Question 6 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.

Find $n$ term and sum to $n$ terms each of the series $28+32+52+152+652+\ldots$

$$\begin{align} & a_2-a_1=32-28=4 \\ & a_3-a_2=52-32=20 \\ & a_4-a_3=152-52=100 \\ & \ldots \quad \cdots \quad \cdots \\ & \cdots \quad \cdots \quad \cdots \\ & a_n-a_{n-1}=(\mathrm{n}-1) \text { term ofthe sequence } 4,20,100, \ldots \end{align}$$ which is a G.P with common ratio $r=5$. Adding column wise, we get $$\begin{align} a_n-a_1&=4+20+100+\ldots+(n-1) \text { terms } \\ & =\dfrac{4[5^{n -1}-1]}{5-1} \\ \Rightarrow a_n \quad a_1&=5^{n-1}-1 \\ \Rightarrow a_n&=5^{n-1}-1+28 \quad \because a_1=28 \\ \Rightarrow a_n&=5^{n-1}+27\end{align}$$ Taking summation of the both sides $$\begin{align}\sum_{r=1}^n a_r&=\sum_{r=1}^n 5^{r-1}+27 \sum_{r=1}^n 1 \\ & =\dfrac{1 \cdot[5^n-1]}{5-1}+27 n \\ \Rightarrow \sum_{r=1}^n a_r&=\dfrac{(5^n-1)}{4}+27 n\\ \text{Hence}\quad5^{n-1}+27; \dfrac{(5^n-1)}{4}+27 n\end{align}$$

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