CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

GEOMETRIC PROGRESSION

Geometric Progression, Its nth term, Sum to n terms, sum to infinite terms, Properties of G.P.,Special Results
GEOMETRIC PROGRESSION
A sequence of non-zero numbers is called a geometric progression (G.P.) if the ratio of the term and the term preceding to it is always a constant quantity.

A sequence a1, a2, a3, a4, ………an, an+1  is called geometric progression.
If  $a_{n+1}=a_{n}\times \: r$ is the common ratio of GP. Where $r=\frac{a_{n+1}}{a_{n}}$
General Geometric Progression is  a, ar, ar2, ar3, ……….., arn-1

First term = a, Second term = ar,  and so on and  r is the common ratio of GP
nth term in GP is =  arn-1
Sum of first n terms in GP is
$S_{n}=\frac{a(r^{n}-1)}{r-1}\: where, r > 1$
Sum of first n terms in GP is
$S_{n}=\frac{a(1-r^{n})}{1-r}\: where,\: \: r < 1$
Sum to infinite terms in GP.
General GP with infinite number of terms is of the following type
ar, ar2, ar3, ……….., arn-1 .........∞
Sum to infinite terms of GP is given by
$S_{\infty }=\frac{a}{1-r}\; , \; Where -1< r< 1 \; or\;\left | r \right | < 1$
SELECTION OF TERMS IN G.P.
$Three\; terms\; in\; \; G.P.\; are:\; \; \; \frac{a}{r},\; a,\; ar, \; \; here \; common \; ratio\; is\; r$
$Four\; terms\; in\; \; G.P.\; are:\; \; \; \frac{a}{r^{3}},\; \frac{a}{r},\; ar, \; ar^{3},\;\; here \; common \; ratio\; is\; r^{2}$$Five\; terms\; in\; \; G.P.\; are:\; \; \; \frac{a}{r^{2}},\; \frac{a}{r}, a,\; ar, \; ar^{2},\;\; here \; common \; ratio\; is\; r$
PROPERTIES OF GEOMETRIC PROGRESSION:-
1) If all the terms of G.P.  be  multiplied  or  divided  by  the  sane  non- zero  constant,    then  it remains a   G.P.   with  common  ratio  r.
2) Reciprocal of the terms of a G.P. form a G.P.
3) If each term of a G.P. raised to the same power , the resulting sequence also form a G.P.
$4)\; If a_{1},\; a_{2},\; a_{3},........\; a_{n}.... are\; the \; terms \; of \; G.P.,\; \; then\;\\ log\; a_{1},log\; a_{2},.....log\; a_{n}........... \; \;\;$$are\; in\; A.P. \; \; and \; \; vice\; -\; versa$
GEOMETRIC MEAN
If a, b, c are the three terms of GP then b is said to be the Geometric Mean and is given by$b=\sqrt{ac}\: \: or\: \: b^{2}=ac$
Explanation:-
If a, b, c are in GP then$\frac{b}{a} = r = \frac{c}{b}$ ⇒ $\frac{b}{a} = \frac{c}{b}$$b^{2}=ac\Rightarrow b=\sqrt{ac}$

Componendo and Dividendo
If four terms a, b, c, d are proportional then $\frac{a}{b}=\frac{c}{d}$
When we apply componendo and dividendo then we get  $\frac{a+b}{a-b}=\frac{c+d}{c-d}$
$or\; \; \frac{Numerator+Denomenator}{Numerator-Denomenator} =\frac{Numerator+ Denomenator}{Numerator-Denomenator}$
SPECIAL RESULTS:-
Sum of first n  natural number  is given by
$1+2+3+4......n=\frac{n(n+1)}{2}$
Sum of square of first n natural number is given by
$1^{2}+2^{2}+3^{2}+......+ n^{2}=\frac{n(n+1)(2n+1)}{6}$
Sum of cube of first n natural number is

$1^{3}+2^{3}+3^{3}+......+n^{3}=\left [ \frac{n(n+1)}{2} \right ]^{2}$

Question:
The first term of an infinite G.P. is 1 and any term is equal to the sum of all terms that follow it. Find the infinite G.P.
Solution:  It is given that:   a = 1

A.T.Q.        Tn = Tn+1 + Tn+2 + Tn+3 +  ……………………

arn-1 = arn + arn+1 + arn+2 + ……………………

$ar^{n-1}=\frac{ar^{n}}{1-r}$Now cross multiply it  and putting a = 1 we get $r^{n-1}-r^{n}=r^{n}\Rightarrow 2r^{n}=r^{n-1}$$\Rightarrow 2r = 1 \Rightarrow r=\frac{1}{2}$

Putting a = 1 and r = 1/2 we get the required sequence as follows $\frac{1}{2},\; \frac{1}{4},\; \frac{1}{16},\; .......,\; \infty$