5.6.2 Progressions, SPM Practice

Question 3:
An arithmetic progression has 16 terms. The sum of the 16 terms is 188, and the sum of the even terms is 96. Find
(a) the first term and the common difference,
(b) the last term.

Solution:
(a) Let the first term = a
Common difference = d

$\begin{array}{l}\text{Given }{S}_{16}=188\\ \text{Thus, }\frac{16}{2}\left[2a+15d\right]=188\\ 8\left[2a+15d\right]=188\\ 2a+15d=\frac{188}{8}\\ 2a+15d=23.5---\left(1\right)\end{array}$

Given the sum of the even terms = 96
$\begin{array}{l}{T}_2+T_4+T_6+\mathrm{\dots ..}+T_{16}=96\\ \left(a+d\right)+\left(a+3d\right)+\left(a+5d\right)+\mathrm{\dots ..}+\left(a+15d\right)=96\\ \frac{8}{2}\left[\left(a+d\right)+\left(a+15d\right)\right]=96\\ 4\left[2a+16d\right]=96\\ 2a+16d=24---\left(2\right)\end{array}$

(2) – (1):
16d – 15d = 24 – 23.5
d = 0.5
Substitute d = 0.5 into (2):
2a + 16 (0.5) = 24
2a + 8 = 24
2a = 16
a = 8
Therefore, first term = 8 and common difference = 0.5.


(b) Last term
= T₂
= 8 + 15 (0.5)
= 8 +7.5
= 15.5

Question 4:
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)
$\begin{array}{l}\text{Given }{T}_3=24\\ a{r}^2=24\mathrm{\dots \dots \dots ..}\left(1\right)\\ \text{Given }{T}_6=7\frac{1}{9}\\ a{r}^5=\frac{64}{9}\mathrm{\dots \dots \dots ..}\left(2\right)\\ \frac{\left(2\right)}{\left(1\right)}:\frac{a{r}^5}{a{r}^2}=\frac{\frac{64}{9}}{24}\\ r^3=\frac{8}{27}\\ r=\frac{2}{3}\end{array}$

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

(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.