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Geometric Progression

August 17, 2020
Category: Geometric Progression

☼ Geometric Progression ( G.P.) : The progression of the form a, ar, ar², ar³, … is known as a G.P., with first term = a and common ratio = r.

In a G.P. a, ar, ar², ar³, . . . , we have :

(i) n^th term, T_n = ar ^{n –}¹.

e.g. (1) Find the 7^th term of 3, 6, 12, . . .

Now, T_n = ar ^{n –}¹

$\therefore T_{7}=3\left (2 \right )^{7-1}$

$= 3\left (2 \right )^{6}$

$=3\left (64 \right )$

$=192$

(2) Find the 6^th term of 2, 10, 50, . . .

( Try yourself ) Ans. : = 6250

(ii) Sum of n^th term,

$S_{n}=\frac{a\left (r^{n}-1 \right )}{r-1}$ : r > 1

$=\frac{a\left (1-r^{n} \right )}{1-r}$ : r< 1

$=na$ : r = 1

$=\frac{1-rT_{n}}{1-r}$ : $r \ne 1$

E.g. (1) Find the sum of first 5 terms of 64, 32, 16, . . .

Now, $S_{n}= \frac{a\left (1-r^{n} \right )}{1-r}$ ; $r=\frac{1}{2} < 1$

$\therefore S_{5}=\frac{64\left (1-\left (\frac{1}{2} \right )^{5} \right )}{1-\frac{1}{2}}$

$=\frac{64\left (1-\ \frac{1}{32} \right )}{\frac{1}{2}}$

$=64\left (\frac{32-1}{32} \right )\times\frac{2}{1}$

$=4(31)$

$=124$

(2) Find the sum of first 8 terms of 3, 6, 12, . . .

( Try yourself. But use the formula $S_{n}= \frac{a\left (1-r^{n} \right )}{1-r}$ ; r = 2 > 1 )

Ans. $S_{8}=765$

(3) Find 2 + 6 + 18 + . . . + 162

Here, a = 2, r = 3 and = 162

Now, $S_{n}=\frac{rT_{n}-a}{r-1}$

$\therefore S_{n}=\frac{3\left (162 \right )-2}{2-1}$

$=\frac{484}{1}$

=484

(4) Find 5 + 10 + 20 + . . . + 160

( Try yourself ) Ans. : = 315

(iii) If a, b, c, d, e, f, . . are in G. P., then

$\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=\frac{e}{d}=\frac{f}{d}=$ common ratio(r)

(iv) If a, b, c are in G. P., then b is called the geometric mean ( G.M.) between a and c. In this case, b = $\sqrt{ac}$

Note : Geometric mean is always a positive number.

e.g. (1) Find the G. M. of 3 and 12.

Now, G.M. = $\sqrt{ac}$ ; where , a=3 and c=12

$=\sqrt{3\times12}$

$=\sqrt{36}$

$=6$

(2) Find the G. M. of 10 and 40.

( Try yourself ) Ans. : 20

(3) Find the G. M. of 3 and 54.

( Try yourself ) Ans. : 9 $\sqrt{2}$

(v) Geometric mean of $a_{1},a_{2},a_{3},...,a_{n},$ is = G. M. = $\sqrt[n]{a_{1}\times a_{2}\times a_{3} \times ...\times a_{n}}$

e.g. (1) Find the G. M. of 2, 8, and 32.

Now, G. M. $=\sqrt[n]{a_{1}\times a_{2}\times a_{3} \times ...\times a_{n}}$

$=\sqrt[3]{2\times8\times 32}$

$= \sqrt[3]{2\times 2^{3}\times 2^{5}}$

$=\sqrt[3]{2^{9}}$

$= 2^{3}$

$=8$

(2) Find the G. M. of 3, 9, 81, 243.

( Try yourself ) Ans. : 27

(vii) If A, , , , . . . , , B are in G. P., then we can say that $a_{1},a_{2},a_{3},...,a_{n},$ are the n geometric means between A and B.

In this case, $r=\left (\frac{b}a{} \right )^{\frac{1}{n+1}}$ and so, $a_{1} = A \times r$

$a_{2} = A \times r^{2}$

. .

$a_{n} = A \times r^{n}$

e.g. (1) Find three geometric means between 2 and 162. or

Fill the blank 2, __, __, __, 162 with appropriate nos. such that the sequence becomes Geometric Sequence.

Now, $r=\left (\frac{b}a{} \right )^{\frac{1}{n+1}}$ ;Where , a = 2 , b = 162 & n = 3

$\therefore r=\left (\frac{162}2{} \right )^{\frac{1}{3+1}}$

$= 82^{\frac{1}{4}} =3$

$\therefore$ Nos, are $2 \times 3=6,$ $6 \times 3=18,$ $18 \times 3=54$

i.e. the blanks are 6, 18, 54.

(2) Find five geometric means between 7 and 448.

( Try yourself ) Ans. : 14, 28, 56, 112, 224.

(vii) It is convenient to take :

For odd number of terms :

Three terms in G. P. are taken as $\frac{a}{r}$ , a, ar

Five terms in G. P. are taken as $\frac{a}{r^{2}}$ , $\frac{a}{r}$ , a, ar, ar

Seven terms in G. P. are taken as $\frac{a}{r^{3}}$ , $\frac{a}{r^{2}}$ , $\frac{a}{r}$ , a, ar, $ar^{2}$ , $ar^{3}$

For even number of terms :

Two terms in G. P. are taken as $\frac{a}{r}$ , ar

Four terms in G. P. are taken as $\frac{a}{r^{3}}$ , $\frac{a}{r}$ , ar, $ar^{3}$

Six terms in G. P. are taken as $\frac{a}{r^{5}}$ , $\frac{a}{r^{3}}$ , $\frac{a}{r}$ , ar, $ar^{3}$ , $ar^{5}$

e.g. (1) If the sum of 3 terms of an arithmetic sequence is 26 and product is 216, then find those terms.

Let, the terms are $\frac{a}{r}, a, ar$

Now,

$\frac{a}{r}+ a+ ar=26$ & $\frac{a}{r}\times a\times ar=216$

$\therefore \frac{6}{r}+6+6r=26$ $\therefore a^{3}=216$

$\therefore \frac{6}{r}-20+6r=0$ $\therefore a=6$

$\therefore \frac{6-20r+6r^{2}}{r}=20$

$\therefore 6r^{2}-20r+6=0$

$\therefore 3r^{2}-10r+3=0$

$\therefore 3r^{2}-9r-r+3=0$

$\therefore \left (3r-1 \right )\left (r-3 \right )=0$

$\therefore r=\frac{1}{3}$ or r=3

$\therefore$ If $r=\frac{1}{3}$ , terms are 18, 6, 2. & If r = 3, terms are 2, 6, 18 $\frac{a}{r}, a, ar$ .

(2) If the sum of 3 terms of an geometric sequence is 32 and product is 400, then find those terms.

( Try yourself ) Ans. : 2, 10, 20 or 20, 10, 2.

Or solve by options directly. It is more easy method.

(viii) The sum of an infinite G.P. $a,ar,ar^2,... is \frac{a}{1-r}$ – 1 < r < 1.

e.g. (1) Find $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$

Here, $r=\frac{1}{2}<1$ and

$\therefore 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...= \frac{1}{1-\frac{1}{2}}$

$=\frac{1}{\frac{1}{2}}$

$=2$

(2) Find $\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+...$

( Try yourself ) Ans. :

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