Question 15 and 16, Exercise 4.3

Solutions of Question 15 and 16 of Exercise 4.3 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.

Find $S_n$ for arithmetic series. $a_{1}=91$, $d=-4$, $a_{n}=15$.

Solution.

Given: $a_{1}=91$, $d=-4$, $a_{n}=15$, $n=?$. We have $$\begin{align} & a_n=a_1+(n-1)d \\ \implies & 15=91+(n-1)(-4) \\ \implies & 15=91-4n+4 \\ \implies & 4n=95-15 \\ \implies & 4n = 80\\ \implies & n = 20. \end{align}$$ Now $$\begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{20}&=\frac{20}{2}[91+15]\\ &=10\times 106\\ &=1060. \end{align}$$ Hence $S_{20}=1060$.

Find $S_n$ for arithmetic series. $d=-4$, $n=9$, $a_{n}=27$.

Solution.

Given: $d=-4$, $n=9$, $a_{n}=27$, $a_1=?$. We have $$\begin{align} & a_n=a_1+(n-1)d \\ \implies & 27=a_1+(9-1)(-4) \\ \implies & 27=a_1-32 \\ \implies & a_1=27+32 =59. \end{align}$$ Now $$\begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{9}&=\frac{9}{2}[59+27]\\ &=\frac{9}{2}\times 86\\ &=387. \end{align}$$ Hence $S_9=387$.

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