Question 2:
The third term and the sixth term of a geometric progression are 24 and $7\frac{1}{9}$ respectively. Find
(a) the first term and the common ratio,
(b) the sum of the first five terms,
(c) the sum of the first n terms with n is very big approaching rⁿ ≈ 0.

Solution:
(a)
$\begin{array}{l}\text{Given }{T}_3=24\\ a{r}^2=24\mathrm{...........}\left(1\right)\\ \text{Given }{T}_6=7\frac{1}{9}\\ a{r}^5=\frac{64}{9}\mathrm{...........}\left(2\right)\\ \frac{\left(2\right)}{\left(1\right)}:\frac{a{r}^5}{a{r}^2}=\frac{\frac{64}{9}}{24}\\ r^3=\frac{8}{27}\\ r=\frac{2}{3}\end{array}$

$\begin{array}{l}\text{Substitute }r=\frac{2}{3}\text{ into }\left(1\right)\\ a{\left(\frac{2}{3}\right)}^2=24\\ a\left(\frac{4}{9}\right)=24\\ a=24\times \frac{9}{4}\\ =54\\ \therefore \text{ the first term 54 and the common ratio is }\frac{2}{3}.\end{array}$

(b)
$\begin{array}{l}{S}_5=\frac{54\left[1-{\left(\frac{2}{3}\right)}^5\right]}{1-\frac{2}{3}}\\ =54\times \frac{211}{243}\times \frac{3}{1}\\ =140\frac{2}{3}\\ \\ \therefore \text{ sum of the first five term is }140\frac{2}{3}.\end{array}$

(c)
$\begin{array}{l}\text{When }-1<r<1\text{ and }n\text{ becomes }\\ \text{very big approaching }{r}^n\approx 0,\\ \therefore S_n=\frac{a}{1-r}\\ =\frac{54}{1-\frac{2}{3}}\\ =162\end{array}$
Therefore, sum of the first n terms with n is very big approaching rⁿ ≈ 0 is 162.

$\begin{array}{l}GP,a=0.8,r=\frac{0.08}{0.8}=0.1\\ S_{\infty }=\frac{a}{1-r}\\ S_{\infty }=\frac{0.8}{1-0.1}\\ S_{\infty }=\frac{0.8}{0.9}\\ S_{\infty }=\frac{8}{9}\to \boxed{\begin{array}{l}\text{check using calculator}\\ \frac{8}{9}=\mathrm{0.888888....}\end{array}}\end{array}$

Solution:
(a)
$\begin{array}{l}2,4,8,\mathrm{.....},8192\leftarrow (\text{Last term is given)}\\ a=2\\ r=\frac{T_2}{T_1}=\frac{4}{2}=2\\ T_n=8192\\ a{r}^{n-1}=8192\leftarrow (Then\text{th term of GP,}{T}_n=a{r}^{n-1})\\ \left(2\right){\left(2\right)}^{n-1}=8192\\ 2^{n-1}=4096\\ 2^{n-1}=2^{12}\\ n-1=12\\ n=13\end{array}$

(b)
$\begin{array}{l}\frac{1}{4},\frac{1}{6},\frac{1}{9},\mathrm{.....},\frac{16}{729}\\ a=\frac{1}{4},r=\frac{\frac{1}{6}}{\frac{1}{4}}=\frac{2}{3}\\ T_n=\frac{16}{729}\\ a{r}^{n-1}=\frac{16}{729}\\ \left(\frac{1}{4}\right){\left(\frac{2}{3}\right)}^{n-1}=\frac{16}{729}\\ {\left(\frac{2}{3}\right)}^{n-1}=\frac{16}{729}\times 4\\ {\left(\frac{2}{3}\right)}^{n-1}=\frac{64}{729}\\ {\left(\frac{2}{3}\right)}^{n-1}={\left(\frac{2}{3}\right)}^6\\ \therefore n-1=6\\ n=7\end{array}$

(c)
$\begin{array}{l}-\frac{1}{2},\text{1},-\text{2},\mathrm{.....},\text{64}\\ a=-\frac{1}{2},r=\frac{-2}{1}=-2\\ T_n=64\\ a{r}^{n-1}=64\\ \left(-\frac{1}{2}\right){\left(-2\right)}^{n-1}=64\\ {\left(-2\right)}^{n-1}=64\times -2\\ {\left(-2\right)}^{n-1}=-128\\ {\left(-2\right)}^{n-1}={\left(-2\right)}^7\\ n-1=7\\ n=8\end{array}$

$\begin{array}{l}\frac{p-20}{p-4}=\frac{p-4}{p+20}\\ \left(p+20\right)\left(p-20\right)=\left(p-4\right)\left(p-4\right)\\ p^2-400=p^2-8p+16\\ 8p=416\\ p=52\end{array}$

Question 13:
An arithmetic progression consists of 10 terms. The sum of the last 5 terms is 5 and the fourth term is 9. Find the sum of this progression.

Solution:

Question 14:
The sum of the first 6 terms of an arithmetic progression is 39 and the sum of the next 6 terms is –69. Find
(a) The first term and the common difference.
(b) The sum of all the terms from the 15th term to the 25th term.

Solution:

Question 15:
An arithmetic progression has 9 terms. The sum of the first four terms is 24 and the sum of all the odd number terms is 55. Find
(a) The first term and common difference,
(b) The seventh term.

Solution:

Question 4:
It is given that -7, h, k, 20,… are the first four terms of an arithmetic progression. Find the value of h and of k.

Solution:

Question 5:
51, 58, 65,... 191 are the first n terms of an arithmetic progression. Find the value of n.

Solution:

Question 6:
Find the number of the multiples of 8 between 100 and 300.

Solution:

Question 15 (2 marks):
It is given that the n^th term of a geometric progression is $T_n=\frac{3r^{n-1}}{2},r\ne k.$  
State
(a) the value of k,
(b) the first term of progression.

Solution:
(a)
k = 0, k = 1 or k = -1 (Any one of these answer).

(b)
$\begin{array}{l}{T}_n=\frac{3}{2}{r}^{n-1}\\ T_1=\frac{3}{2}{r}^{1-1}\\ =\frac{3}{2}{r}^0\\ =\frac{3}{2}\left(1\right)\\ =\frac{3}{2}\end{array}$

Question 16 (4 marks):
It is given that p, 2 and q are the first three terms of a geometric progression.
Express in terms of q
(a) the first term and the common ratio of the progression.
(b) the sum to infinity of the progression.

Solution:
(a)
$\begin{array}{l}{T}_1=p,T_2=2,T_3=q\\ \frac{T_2}{T_1}=\frac{T_3}{T_2}\\ \frac{2}{p}=\frac{q}{2}\\ p=\frac{4}{q}\\ \mathrm{First}\text{ term},T_1=p=\frac{4}{q}\\ \text{Common ratio}=\frac{q}{2}\end{array}$

(b)
$\begin{array}{l}a=\frac{4}{q},r=\frac{q}{2}\\ S_{\infty }=\frac{a}{1-r}\\ =\frac{\frac{4}{q}}{1-\frac{q}{2}}\\ =\frac{4}{q}÷\left[1-\frac{q}{2}\right]\\ =\frac{4}{q}÷\left[\frac{2-q}{2}\right]\\ =\frac{4}{q}\times \frac{2}{2-q}\\ =\frac{8}{2q-q^2}\end{array}$

Question 6 (7 marks):
Diagram shows part of a rectangular wall painted with green, G, blue, B and purple, P subsequently.
The height of the wall is 2 m. The side length of the first coloured rectangle is 5 cm and the side length of each subsequent coloured rectangle increases by 3 cm.


It is given that the total number of the coloured rectangles is 54.
(a) Find
(i) the side length, in cm, of the last coloured rectangle,
(ii) the total length, in cm, of the painted wall.
(b) Which coloured rectangle has an area of 28000 cm²?
  Hence, state the colour of that particular rectangle.

Solution:
(a)
5, 8, 11, …
a = 5, d = 3

(i)
T₅₄ = 1 + (54 – 1)d
= 5 + 53(3)
=164 cm

(ii)
$\begin{array}{l}{S}_n=\frac{n}{2}\left(a+l\right)\\ S_{54}=\frac{54}{2}\left(5+164\right)\\ =4563\text{ cm}\end{array}$


(b)
Area of the first rectangle
= 2 m × 5 cm
= 200 × 5
= 1000 cm

Area of the second rectangle
= 200 × (5 + 3)
= 1600 cm

Area of the third rectangle
= 200 × (5 + 3 + 3)
= 2200 cm

1000, 1600, 2200, …
a = 1000, d = 600
T_n = 28 000
a + (n – 1)d = 28 000
1000 + (n – 1)600 = 28 000
600(n – 1) = 27 000
n – 1 = 45
n = 46

The colour of that particular rectangle is green.

Question 5 (6 marks):
The sum of the first n terms of an arithmetic progression, S_n is given by $S_n=\frac{3n\left(n-33\right)}{2}.$  
Find
(a) the sum of the first 10 terms,
(b) the first term and the common difference,
(c) the value of q, given that q^th term is the first positive term of the progression.

Solution:
(a)
$\begin{array}{l}{S}_n=\frac{3n\left(n-33\right)}{2}\\ S_{10}=\frac{3\left(10\right)\left(10-33\right)}{2}\\ S_{10}=-345\end{array}$

(b)
$\begin{array}{l}{S}_n=\frac{3n\left(n-33\right)}{2}\\ S_1=\frac{3\left(1\right)\left(1-33\right)}{2}\\ S_1=-48\\ T_1=S_1=-48\\ \\ \text{First term, }a=T_1=-48\\ T_n=S_n-S_{n-1}\\ T_2=S_2-S_1\\ T_2=\frac{3\left(2\right)\left(2-33\right)}{2}-\left(-48\right)\\ T_2=-45\\ \\ \text{Common difference, }d\\ =T_2-T_1\\ =-45-\left(-48\right)\\ =3\end{array}$

(c)
$\begin{array}{l}\text{First positive term, }{T}_q>0\\ T_q>0\\ a+\left(q-1\right)d>0\\ -48+\left(q-1\right)3>0\\ -48+3q-3>0\\ 3q>51\\ q>17\\ \text{Thus, }q=18.\end{array}$

Question 13:
The sum of the first n term of a geometry progression is given by 6(1 – 0.5ⁿ ). Find
(a) The fourth term of the progression,
(b) The sum to infinity of the progression.

Solution:

Question 14:
A gardener has the task of digging an area of 800 m². On the first day he digs an area of 10 m². On each successive day he digs an area of 1.2 times the area that he dug the previous day, until the day when the task is completed. Find the number of days needed to complete the task.

Solution: