A geometric progression is a sequence in which the consecutive terms differ by a constant factor.

The said constant factor is called the

**common ratio. "r"**is used to denote the common ratio.

As with arithmetic sequence, the first term of a geometric progression is denoted by the letter "

**a**".

__Example 1__Find the first term and common ratio of the G.P 3, 12,48,......

**Solution:**

The first term is already given as 3

To find the common ratio, we divide consecutively the terms of the G.P, that is $\frac{T_2}{T_1}$

$\frac{12}{3}=4$

Therefore, the first term and common ratio are 3 and 4 respectively

__The nth term of a G.P__Generally, the nth term of a G.P is given as:

$T_n=ar^{n-1}$

With this fact, let's solve some examples

__Example 2__Find the fifth term of a Geometric progression which has a first term of 18 and a common ratio of ⅓.

**Solution:**

Using the formula G.P

$T_5=18(\frac{1}{3})^{5-1}$

$T_5=18(\frac{1}{3})^4$

$T_5=18(\frac{1}{81})$

$T_5=\frac{18}{81}$

$T_5=\frac{2}{9}$

__Example 3__The first term of a G.P is 27. Its common ratio is ⅔. Find the fifth term?

**Solution:**

$T_5=27(\frac{2}{3})^{5-1}$

$T_5=27(\frac{2}{3})^4$

$T_5=27(\frac{16}{81})$

$T_5=\frac{16}{3}$

__Example 4__If the fourth term of a G.P which has a common ratio of 3 is 189. Find the first term of the G.P.

**Solution:**

$189=a(3^{4-1})$

$189=a(3^3)$

$189=27a$

$\frac{189}{27}=\frac{27a}{27}$

$a=7$

__Sum of the nth term of a G.P__Generally, the sum of a G.P is given as:

$S_n=\frac{a(1-r^n)}{1-r}$ when r<1

$S_n=\frac{a(r^n-1)}{r-1}$ when r>1

The first formula is used when the common ratio is lesser than one.

The second formula is used when the common ratio is greater than one.

__Example 5__Find the sum of the first four terms of a G.P if its first term and common ratio are 153 and ⅓ respectively.

**Solution:**

Here, the common ratio is lesser than one. Hence, the first formula is used

$S_5=\frac{153(1-(\frac{1}{3})^4)}{1-\frac{1}{3}}$

$S_5=\frac{153(1-\frac{1}{81}}{1-\frac{1}{3}}$

$S_5=\frac{153(\frac{81-1}{81})}{\frac{3-1}{3}}$

$S_5=\frac{\frac{12240}{81}}{\frac{2}{3}}$

$S_5=\frac{12240}{81}÷\frac{2}{3}$

Now, we invert the fraction and change ÷ to ×

$S_5=\frac{12240}{81}\times\frac{3}{2}$

$S_5=\frac{36720}{162}$

$S_5=\frac{680}{3}$

__Example 6__The first term of a G.P is 7. its common ratio is 5. Find the sum of the first five terms of the G.P.

**Solution:**

The common ratio is greater than one, hence, the first formula is used.

$S_5=\frac{7(5^5-1)}{5-1}$

$S_5=\frac{7(3125-1)}{4}$

$S_5=\frac{7(3124)}{4}$

$S_5=\frac{7(781)}{1}$

$S_5=5467$

__Related posts__

__Example 7__Find the sum of the first four terms of a G.P whose first and second terms are 5 and 10 respectively.

**Solution:**

To derive the common ratio, we divide consecutive terms, hence,

$\frac{10}{5}=2$

The common ratio is greater than one, therefore, the first formula is used.

$S_4=\frac{5(2^4-1)}{2-1}$

$S_4=5(16-1)$

$S_4=5(15)$

$S_4=75$

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