Question 10:
The sequence –11, –5, 1,… is an arithmetic progression. State the three consecutive terms of this arithmetic progression where the sum of these three terms is 93.

Solution:

Question 11:
An arithmetic series, with the first term 12 and common difference d, consists of 23 terms. Given that the sum of the last 3 terms is 5 times the sum of the first 3 terms, find
(a) the value of d,
(b) the sum of the first 19 terms.

Solution:

Question 12:
The sum of n terms of an arithmetic progression is given by the formula $S_n=\frac{n}{2}(5-3n)$ . Find
(a) the first term,
(b) the common difference,
(c) the tenth term.

Solution:

Question 7:
Find the sum of all the multiples of 7 between 100 and 500.

Solution:

Question 8:
If ${\log }_{10}p,{\log }_{10}pq\text{ and }{\log }_{10}p{q}^2$ are the first three terms of a progression, show that it forms an arithmetic progression.

Solution:

Question 9:


Show that the volumes of the cylinders in the above diagram form an arithmetic progression and state its common difference.

Solution:

Question 1:
The third and eighth terms of an arithmetic progression are –5 and 15 respectively. Find
(a) The first term and the common difference
(b) The sum of the first 10 terms.

Solution:
(a)
$\begin{array}{l}{T}_3=-5\to \text{ Use }{T}_n=a+\left(n-1\right)d\\ a+2d=-5\to \left(1\right)\\ T_8=15\\ a+7d=15\to \left(2\right)\\ \\ \left(2\right)-\left(1\right),\\ 5d=20\\ d=4\\ \text{Substitute }d=4\text{ into }\left(1\right),\\ a+2\left(4\right)=-5\\ a=-13\end{array}$

(b)
$\begin{array}{l}{S}_{10}=\frac{10}{2}\left[2\left(-13\right)+9\left(4\right)\right]\to \text{Use }{S}_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\\ =50\end{array}$

Question 2:
The first three terms of an arithmetic progression are 2k, 3k + 3, 5k + 1. Find
(a) the value of k,
(b) the sum of the first 15 terms of the progression.

Solution:

Question 3:
Given an arithmetic progression p + 9, 2p + 10, 7p – 1,… where p is a constant. Find
(a) value of p,
(b) the sum of the next five terms.

Solution:

Tn = a + (n − 1) d where a = first term d = common difference n = the number of term Tn  = the nth term

If a, b, c are three consecutive terms of an arithmetic progression, then c – b = b – a

$\begin{array}{l}x=\frac{1}{3}\\ \frac{1}{3}+1,\text{2}\left(\frac{1}{3}\right)+3,6\\ \frac{\text{4}}{\text{3}}\text{, 3}\frac{2}{3}\text{, 6}\\ d=\text{3}\frac{2}{3}-\frac{\text{4}}{\text{3}}=2\frac{1}{3}\end{array}$

Solution:

$\begin{array}{l}{T}_n=a{r}^{n-1}\\ T_1=a{r}^{1-1}=a{r}^0=a\leftarrow (\text{First term})\\ T_2=a{r}^{2-1}=a{r}^1=ar\leftarrow (Second\text{term})\\ T_3=a{r}^{3-1}=a{r}^2\leftarrow (Third\text{term})\\ T_4=a{r}^{4-1}=a{r}^3\leftarrow (\text{Fourth term})\end{array}$

(a)
$\begin{array}{l}8,4,2,\mathrm{.....}\\ a=8,r=\frac{4}{8}=\frac{1}{2}\\ T_8=a{r}^7\\ T_8=8{\left(\frac{1}{2}\right)}^7=\frac{1}{16}\end{array}$

(b)
$\begin{array}{l}\frac{16}{27},\frac{8}{9},\frac{4}{3},\mathrm{.....}\\ a=\frac{16}{27}\\ r=\frac{T_2}{T_1}=\frac{\frac{16}{27}}{\frac{8}{9}}=\frac{2}{3}\\ T_6=a{r}^5\\ =\frac{16}{27}{\left(\frac{2}{3}\right)}^5=\frac{512}{6561}\end{array}$

(a)
$\begin{array}{l}-11,-8,-5,\mathrm{.....}\text{Find}{S}_{15}\\ a=-11,\\ d=-8-\left(-11\right)=3\\ S_{15}=\frac{15}{2}\left[2a+14d\right]\\ S_{15}=\frac{15}{2}\left[2\left(-11\right)+14\left(3\right)\right]=150\end{array}$

(b)
$\begin{array}{l}8,10\frac{1}{2},13,\mathrm{.....}\text{Find}{S}_{13}\\ a=8\\ d=10\frac{1}{2}-8=\frac{5}{2}\\ S_{13}=\frac{13}{2}\left[2a+12d\right]\\ S_{13}=\frac{13}{2}\left[2\left(8\right)+12\left(\frac{5}{2}\right)\right]=299\end{array}$

(c)
$\begin{array}{l}5,7,9,\mathrm{.....},75\leftarrow \left(\text{The last term}l=75\right)\\ a=5\\ d=7-5=2\\ S_n=\frac{n}{2}\left(a+l\right)\\ S_{36}=\frac{36}{2}\left(5+75\right)=1440\\ \\ \text{The last term}l=75\\ T_n=75\\ a+\left(n-1\right)d=75\\ 5+\left(n-1\right)\left(2\right)=75\\ \left(n-1\right)\left(2\right)=70\\ n-1=35\\ n=36\end{array}$

Question 1:
The fourth term of a geometric progression is –20. The sum of the fourth and the fifth term is –16. Find the first term and the common ratio of the progression.

Solution:

Question 2:
The fourth and the seventh terms of a geometric progression are 18 and 486 respectively. Find the third term.

Solution:

Question 3:
For a geometric progression, the sum of the first two terms is 30 and the third term exceeds the first term by 15. Find the common ratio and the first term of the geometry progression.

Solution:
$\begin{array}{l}{T}_1+T_2=30\\ a+ar=30\\ a\left(1+r\right)=30---(1)\\ T_3-T_1=15\\ a{r}^2-a=15\\ a\left(r^2-1\right)=15---(2)\\ \\ \frac{\left(2\right)}{\left(1\right)}=\frac{a\left(r^2-1\right)}{a\left(1+r\right)}=\frac{15}{30}\\ \frac{\left(r-1\right)\left(r+1\right)}{1+r}=\frac{1}{2}\to \boxed{\begin{array}{l}\left(r^2-1\right)=\\ \left(r-1\right)\left(r+1\right)\end{array}}\\ r-1=\frac{1}{2}\\ r=\frac{1}{2}+1=\frac{3}{2}\\ \\ \text{From (1),}\\ a\left(1+r\right)=30\\ a\left(1+\frac{3}{2}\right)=30\\ \frac{5a}{2}=30\\ a=12\end{array}$

Smart TIPS: Solving the simultaneous equation of a and r. Using the formula T _n = a r ⁿ⁻¹

Solution:
$\begin{array}{l}{T}_6=32\\ a{r}^5=32\text{ ----- (1)}\\ T_3=4\\ a{r}^2=4\text{ ----- (2)}\\ \frac{(1)}{(2)}=\frac{a{r}^5}{a{r}^2}=\frac{32}{4}\\ r^3=8\\ r=2\\ \\ \text{Substitute }r=2\text{ into (2),}\\ a{\left(2\right)}^2=4\\ a=1\end{array}$

Solution:
$$\begin{array}{l}\mathrm{}\end{array}$$

Solution:
$\begin{array}{l}3,12,48,\mathrm{.....}\\ GP,a=3,r=\frac{T_2}{T_1}=\frac{12}{3}=4\\ T_n>1000000\\ \left(3\right){\left(4\right)}^{n-1}>1000000\\ {\left(4\right)}^{n-1}>\frac{1000000}{3}\leftarrow \left[\left(3\right){\left(4\right)}^{n-1}\ne {12}^{n-1}\right]\\ \log 4^{n-1}>\log \frac{1000000}{3}\leftarrow (\text{Put log for both sides)}\\ \left(n-1\right)\lg 4>\lg \frac{1000000}{3}\leftarrow \left({\log }_a{m}^n=n{\log }_{a}m\right)\\ \left(n-1\right)\left(0.6021\right)>5.523\\ n-1>9.17\\ n>10.17\\ n=11\leftarrow (n\text{is an integer)}\\ \\ \boxed{\begin{array}{l}Check:\\ T_{11}=(3){\left(4\right)}^{10}\\ T_{11}=3145728>1000000\end{array}}\end{array}$