Question 11, 12 and 13, Exercise 4.5

Solutions of Question 11, 12 and 13 of Exercise 4.5 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 $a_{1}$ for the geometric series: $S_{n}=244, r=-3, n=5$

Solution.

Given: $S_{n}=244$, $r=-3$, $n=5$.
We know $$S_n =\frac{a_1(1-r^n)}{1-r}, \quad r\neq 1.$$ Thus $$\begin{align*} & 244=\frac{a_1(1-(-3)^5)}{1-(-3)} \\ \implies & 244=\frac{a_1(1+243)}{4} \\ \implies & 976=244a_1\\ \implies & a_1=4.\\ \end{align*}$$ Hence $a_1=4$.

Find $a_{1}$ for the geometric series: $S_{n}=32, r=2, n=6$

Solution.

Given: $S_{n}=32$, $r=2$, $n=6$.
We know $$S_n = \frac{a_1(1-r^n)}{1-r}, \quad r \neq 1.$$ Thus, $$\begin{align*} 32 &= \frac{a_1(1-2^6)}{1-2} \\ &= \frac{a_1(1-64)}{-1} \\ &= \frac{a_1(-63)}{-1} \\ &= 63a_1 \\ \implies a_1 &= \frac{32}{63}\\ \implies a_1&= 0.51 \end{align*}$$ Hence, $a_1 = 0.51$

Find $a_{1}$ for the geometric series: $a_{n}=324, r=3, S_{n}=484$

Solution.

Given: $a_{n}=324$, $r=3$, $S_{n}=484$.

As we know $$S_n =\frac{a_1-a_n r}{1-r}, \quad r\neq 1.$$ Thus $$\begin{align*} & 484=\frac{a_1-(324)(3)}{1-3} \\ \implies & 484=\frac{a_1-972}{-2} \\ \implies & a_1-972=-968 \\ \implies & a_1=-968+972 \\ \implies & a_1=4 \\ \end{align*}$$ Hence $a_1=4$.

< Question 9 & 10 Question 14 >