Progressions Complete Tutorial - II - CAT 2009 Online Preparation

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CAT 2009 Online Preparation

Progressions Complete Tutorial - II

Author: Sureshbala

GEOMETRIC PROGRESSION (G.P.)

Quantities are said to be in Geometric Progression when the ratio of any quantity to the quantity that follows it is the same. In other words, any term of a G.P. can be obtained by multiplying the previous term by a constant factor.

This constant factor is called the common ratio (and is normally represented by r). The first term of a G.P. is obtained by a.

A G.P. can be represented by a, ar, ar^2 ,..... where a is the first term and r is the common ratio of the G.P. $n^{th}$ term of G.P.is $ar^{n-1}$

$\text{Sum to n terms} = \cfrac{a(1-r^n)}{1-r} \; or \; \cfrac{a(r^n -1)}{1-r}$
$\text{This can also be written as} \; \cfrac{a(r^n - 1)}{1 - r}$

$= \cfrac{rar^{n-1}-a}{r-1} = \cfrac{r \times Last \; term - First \; term}{r-1}$

Thus the sum to n terms of a geometric progression can also be written as

$S_n = \cfrac{r \times Last \; Term - First \; term}{r-1}$

If n terms $a_1, a_2, a_3 \cdot{.... ... ...} a_n$ are in G.P., then the Geometric Mean (G.M.) of these n terms is given
by $n\sqrt{a_1 \times a_2 \times a_3 \times \cdots \times a_n}$

If three terms are in geometric progression then the middle term is a geometric Mean of the other two terms, i.e., if a, b and c are in G.P., then b is the geometric mean of the three terms and b^2 = ac

If there are two terms a and b, their geometric mean is given by $G.M. = \sqrt{ab}$

For any two unequal positive numbers a and b, their Arithmetic Mean is always greater than their Geometric Mean, i.e.,

For any two unequal positive number a and b
$\cfrac{a+b}{2} > \sqrt{ab}$

When there are three terms in geometric progression, we can take the three terms to be a/r, a and ar

INFINATE GEOMETRIC PROGRESSION

If -1 < r < +1 or |r| <1, then sum of a geometric progression does not increase infinitely but "converges" to a particular point. Such a G.P is referred to as an infinite geometric progression. The sum of an infinite geometric progression is represented by $S_\alpha$ and is given by the formula.

$S_\alpha = \cfrac{\alpha}{1-r}$

SOME IMPORTANT RESULTS

The sum to n terms of the following series are quite useful and hence should be remembered by students. Sum of the first n natural numbers

$\Sigma n = \cfrac{n(n+1)}{2}$

Sum of squares of the first n natural numbers

$\Sigma n^3 = \Big [ \cfrac{n(n+1)}{2}\Big ]^2 = \cfrac{n^2 (n+1)^2}{4} = [\Sigma n]^2$

Example

1. Find the $7^{th}$ terms of the G.P., whose $1^{st}$ term is 6 and common ratio is 2/3.

Sol . $n^{th} term = a.r^n - 1; 7^{th} term = 6. \bigg(\cfrac{2}{3}\bigg)^6$

$=\cfrac{6 \times 64}{729} = \cfrac{384}{729} = \cfrac{128}{243}$

2. Find the sum to 5 terms of a G.P., whose $1^{st}$ term is 16 and the common ratio is 1/2

Sol.
$\text{Sum to 5 terms} = \cfrac{a(1-r^n)}{(1-r)}$
$=\cfrac {16\{1-(\frac {1}{2})^5\}}{(1-\frac {1}{2})}$ $=\cfrac{16(1-\frac{1}{32})}{(1-\frac{1}{2})}$ $=16*\cfrac{31}{32}* \cfrac{2}{1} = 31$

3. Find the common ratio of the G.P. whose first and last term are 25 and 1/625 and the sum of the G.P is 19531/625.

Sol. We know sum of a G.P is

$\cfrac{r - last \; term - first \; term }{r-1}$

Since we can see the last term in less than the first term for the sake of convenience, we write the sum as

$\cfrac{first \; term - last \; term}{1-r} \Rightarrow \cfrac{19531}{625} = \cfrac{25- \frac{r}{625}}{1-r}$

On simplification, this yields r = 1/5

4. Find the number of terms in the G.P whose $1^{st}$ term is 16, sum is 1365/64 and common ratio is (1/4).

Sol.
$\text{Sum of G.P} = \cfrac{First \; term - last \; term}{1-r}$ $\Rightarrow \cfrac{1365}{64} = \cfrac{16-1/4(last \; term)}{3/4}$
$\Rightarrow \cfrac{4095}{256} = 16 - \cfrac{1}{4} (last \; term)$
$\Rightarrow \cfrac{1}{4}(last \; term) = \cfrac{1}{256}$
$\Rightarrow last \; term = \cfrac{1}{64}= \cfrac{16}{1064} = 16.(\cfrac{1}{1024})$
$\text{last term} = 16 \bigg(\cfrac{1}{4} \bigg)^5$

This is of the form . Hence $a.r^{n-1} = 6$ hence n = 6

$\therefore \text{Number of terms in the G.P is 6.}$

5. Find three numbers in G.P whose sum is 26 and product is 216.

Sol. Let the 3 numbers be a/r, a and ar.

Given that

a/r. a. ar = 216;

$\Rightarrow a^3 = 216; a=6$

a/r + a + ar = 26

$\Rightarrow 6 + 6r + 6r^2 = 26r$

6 r^2 - 20 r + 6 = 0

$\Rightarrow 6r(r-3) - 2 (r-3) = 0$

$\Rightarrow r = \cfrac{1}{3} \; (or) \; r =3$

Hence the three numbers are 2, 6 and 18.

Note : Even if the other value of r is taken values of numbers will be same.

6. The sum of an infinite G.P is 18 and the sum of their squares is 162. Find the series.

Sol: Let $1^{st}$ term be a and common ratio be r

$\therefore$ Some of the infinite terms = $\cfrac{a}{1-r}$ and

Sum of the Squre = $\cfrac{a^2}{1-r^2}$

$\cfrac{a}{1-r} = 18 \And \cfrac{a^2}{1-r^2} = 162 \Rightarrow \cfrac{a}{1+r} = 9$

$\therefore \cfrac{1+r}{1-r} = 2 \Rightarrow 1+r = 2 - 2r \Rightarrow r = \cfrac{1}{3}$
$\therefore a = 12$

Hence the series is 12, 4, 4/3, 4/9,……

7. If |x| < 1, find the sum of the series $2 + 4x + 6x^2 + 8x^3 + \cdots$

Sol. Let $S = 2 + 4x + 6x^2 + 8x^3 + \cdots (1)$

$\cfrac{xS = 2x + 4x^2 + 6x^3 + \cdots (2)}{(1)-(2) \; gives}$

$S(1-x) = 2 + 2x + 2x^2 + 2x^3 + \cdots$
$=2 (1 + x + x^2 + \cdots)$
$= 1 + x + x^2 + \cdots$ is an infinite G.P with a = 1 and

r = x, and as |x| < 1, |r| < 1. Hence,

$\therefore \text{Sum of thies series} = \cfrac{1}{1- x}$

$\therefore S(1-x) = \cfrac{2}{1-x}$
$\therefore S = \cfrac{2}{(1-x)^2}$

8. Find the sum of the series $1, 2/5, 4/25, 8/125,\cdots \infty$

Sol: By observation, we can make out this is an infinite G.P with a = 1 and r = 2/5.

$\therefore \; \text{Sum of the G.P} = \cfrac{a}{1-r} = \cfrac{1}{1-(\frac{2}{5})} = \cfrac{5}{3}$

9.Find the sum of series $3\sqrt{3}, 9\sqrt{5}, 9\sqrt{\cfrac{3}{5}} \cdots \infty$

Sol: here, the ratio of the $2^{nd}$ and the $1^{st}$ term is $\sqrt{\cfrac{3}{5}}$

The ratio of the $3^{rd}$ and the $2^{nd}$ ratio is also $\sqrt{3/5}$

$\therefore$ This is a G.P. with $a = 3\sqrt{3}$ and $r = \sqrt{\frac{3}{5}}$

$\therefore$ $Sum = \cfrac{a}{1-r}$

$= \cfrac{3\sqrt{3}}{1- \frac{\sqrt{3}}{\sqrt{5}}}$ $=\cfrac{3\sqrt{15}}{\sqrt{5} - \sqrt{3}} \times \cfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} +\sqrt{3}}$

$= \cfrac{3|\sqrt{75} + \sqrt{45}|}{2} = \cfrac{3(5\sqrt{3} + 3\sqrt{5)}}{2}$

10 if (1^3 - t_1) + (2^3 - t_2) + $(3^3 - t_3) + (4^4 - t_4) + \cdots + (n^3 - t_n)$

$= \cfrac{n^2 (n-3)}{4}$ , find t_n

where $t_1,t_2, t_3,\cdots t_n$ are the terms of a series

Sol :

$(1^3 + 2^3 + 3^3 + \cdots + n^3) - (t_1, t_2 + \cdots + t_n)$
$=\cfrac{n^2 (n-3)}{4}$
$\Rightarrow S_n = \cfrac{n^2(n+1)^2}{4} - \cfrac{1}{4} n^2 (n-3)$
$\Rightarrow \cfrac{n^4 + 2n^3 + n^2 - n^3 + 3n^2}{4}$
$\Rightarrow \cfrac{n^4 + n^3 + 4n^2}{4}$
$S_n - 1 = (n-1)^2 \{ (n-1)^2 + (n-1) + 4\}$
$= \cfrac{(n^2 - 2n + 1)(n^2 - n + 4)}{4} = \cfrac{n^4 - 3n^3 + 7n^2 - 9n + 4}{4}$
t_n = S_n - S_n -1