11th Class Mathematics Sequence and Series Geometric Progression

      Geometric Progression (G.P.)

A sequence is said to be in G.P. if the ratio between the consecutive terms is constant. The sequence \[{{a}_{1}},{{a}_{2}},{{a}_{3}},---,{{a}_{n}}\] is said to be in G.P. if the ratio of the consecutive term is a constant.

If 'r' is the common ratio, then the nth term of the sequence is given by \[{{a}_{n}}=a\,\,{{r}^{n-1}}\]

The sum of n terms of the G.P. sequence is given by

\[{{S}_{n}}=\frac{a({{r}^{n}}-1)}{r-1}If\,\,r\,>\,1\,and\,{{S}_{n}}=\frac{a(1-{{r}^{n}})}{1-r}if\,r\,>1\]

Sum to infinity is a G.P. series is given by \[{{S}_{\propto }}=\frac{a}{1-r}.\]  

Geometric Mean (G.M.)

If 'a' and 'b' are any two terms of G.P., then the geometric mean is given by \[GM=\sqrt{ab}\]  

Properties of GP

(a) If each term of GP is multiplied or divided by a constant, then the resulting sequence is also in GP.

(b) If each term of the GP is raised to the same power then the resulting sequence is also in GP.

(c) The reciprocal of each term of GP also results in GP.  

The sum of the series of the sequence given by \[\frac{2}{3},-1,\frac{3}{2},.....\]. to 5 terms is given by:

(a) 0                                                         

(b) \[\frac{55}{24}\]

(c) \[\frac{65}{34}\]                                                           

(d) \[\frac{65}{24}\]

(e) None of these

Answer: (b)

Explanation

Since the above sequence is in GP, and first term is \[a=\frac{2}{3}\] and common ratio \[=r=-\frac{3}{2}\]

The sum of n terms of GP is given by \[{{S}_{n}}=\frac{a(1-{{r}^{n}})}{1-r}\]

\[\Rightarrow \,\,{{S}_{n}}\frac{\frac{2}{3}\left\{ 1-{{\left( -\frac{3}{2} \right)}^{5}} \right\}}{1-\left( -\frac{3}{2} \right)}\]

\[\Rightarrow \,\,{{S}_{n}}\frac{4}{15}\left\{ 1+\frac{243}{32} \right\}\]

\[\Rightarrow \,\,{{S}_{n}}=\frac{55}{24}\]

Robert and James were playing a game and Robert asks James to find the three numbers which are in GP such that their sum is 19 and their product is 216. The three numbers are.

(a) (4, 6, 9)                                             

(b) (3, 6, 12)

(c) (5, 10, 20)                                        

(d) (7, 14, 28)

(e) None of these              

Answer: (a)

Explanation

Let 'a' be the first term of the GP and V be the common ratio. Then the three terms which are in GP can be written\[as\frac{a}{r},a,ar.\].

According to question,

\[\frac{a}{r}\times a\times ar=21\]

\[\Rightarrow \,\,{{a}^{3}}=216\]

\[\Rightarrow \,\,a=6\]

Also, \[\frac{a}{r}+a+ar=19\]

\[\Rightarrow \,\,\frac{6}{r}+6++6r=19\]

\[\Rightarrow \,\,6{{r}^{2}}-13r+6=0\]

\[\Rightarrow \,\,\frac{3}{2}or\frac{2}{3}\]

There is a sequence of numbers such that they are in GP and sum to infinite of the number of terms of the sequence is 15 and the sum of their square is 45. The first term of the sequence is given by:

(a) 3                                                         

(b) 5

(c) 9                                                          

(d) 13

(e) None of these

Answer: (b)

Explanation

The sum of infinite number of terms in GP is given by \[{{S}_{\propto }}=\frac{a}{1-r}\]

\[\Rightarrow \,\,15=\frac{a}{1-{{r}^{2}}}----(1)\]

Also the sum of square of the terms of the GP is given by,

\[{{S}_{\propto }}=\frac{{{a}^{2}}}{1-{{r}^{2}}}\]

\[\Rightarrow \,\,45=\frac{{{a}^{2}}}{1-{{r}^{2}}}----(2)\]

Solving (1) and (2) we get,

a = 5

The three numbers in GP are such that their continued product is 216, and the sum of product in pairs is 156; then the three numbers are:

(a) (2, 4, 8)                                             

(b) (1, 5, 25)

(c) (3, 9, 27)                                           

(d) (2, 6, 18)

(e) None of these

Answer: (d)

If the three terms x, y, z are the \[{{p}^{th}},{{q}^{th}}\,and\,{{r}^{th}}\] terms of GP. Then the value of\[{{X}^{q-r}}{{Y}^{r-p}}{{Z}^{p-q}}\].

(a) 1                                                         

(b) 0

(c) 3                                                          

(d) 5

(e) None of these

Answer: (a)