Geometric progression

EXERCISE 20.2

QUESTION 1

Find three numbers in G.P. whose sum is 65 and whose product is 3375 .

Sol :

Let the terms of the the given G.P. be \frac{a}{r}, a and ar

Then, product of the G.P. =3375

\Rightarrow a^{3}=3375

a = 15

Similarly, sum of the G.P. =65

\Rightarrow \frac{a}{r}+a+a r=65

Substituting the value of a

\frac{15}{r}+15+15 r=65

\Rightarrow 15 r^{2}+15 r+15=65 r

\Rightarrow 15 r^{2}-50 r+15=0

\Rightarrow 5\left(3 r^{2}-10 r+3\right)=0

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

\Rightarrow(3 r-1)(r-3)=0

\Rightarrow r=\frac{1}{3}, 3

Hence, the G.P.for a=15 and r=\frac{1}{3} is 45 , 15 , 5

And, the G.P. for a=15 and r=3 is 5 , 15 , 45 

 

QUESTION 2

Find three numbers in G.P. whose sum is 38 and their product is 1728 

Sol :

Let the terms of the the given G.P.be \frac{a}{r}, a and ar

Then, product of the G.P. = 1728

\Rightarrow a^{3}=1728

a = 12

Similarly, sum of the G.P. = 38

\Rightarrow \frac{a}{r}+a+a r=38

Substituting the value of a

\frac{12}{r}+12+12 r=38

\Rightarrow 12 r^{2}+12 r+12=38 r

\Rightarrow 12 r^{2}-26 r+12=0

\Rightarrow 2\left(6 r^{2}-13 r+6\right)=0

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

\Rightarrow r=\frac{2}{3}, \frac{3}{2}

Hence, the G.P. for a=12 and r=\frac{2}{3} is 18,12 and 8 .

And, the G.P. for a=12 and r=\frac{3}{2} is 8,12 and 18

Hence, the three numbers are 8 , 12 and 18

 

QUESTION 3

The sum of first three terms of a G.P.is 13/ 12 and their product is -1 Find the G.P.

Sol :

Let the first three numbers of the given G.P. be \frac{a}{r}, a and ar

Product of the G.P. = -1

\Rightarrow a^{3}=-1

a = -1

Similarly, Sum of the G.P. =\frac{13}{12}

\Rightarrow \frac{a}{r}+a+a r=\frac{13}{12}

Substituting the value of a=-1

\frac{-1}{r}-1-r=\frac{13}{12}

\Rightarrow 12 r^{2}+25 r+12=0

\Rightarrow 12 r^{2}+16 r+9 r+12=0

\Rightarrow 4 r(3 r+4)+3(3 r+4)=0

\Rightarrow(4 r+3)(3 r+4)=0

\Rightarrow r=-\frac{3}{4},-\frac{4}{3}

Hence, the G.P. for a=-1 and r=-\frac{3}{4} is \frac{4}{3},-1 and \frac{3}{4}

And, the G.P. for a=-1 and r=-\frac{4}{3} is \frac{3}{4},-1 and \frac{4}{3}

 

QUESTION 4

The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is 871/ 2 Find them.

Sol :

Let the required numbers be \frac{a}{r}, a and ar

Product of the G.P. = 125

\Rightarrow a^{3}=125

a = 5

Sum of the products in pairs =87 \frac{1}{2}=\frac{175}{2}

\Rightarrow \frac{a}{r} \times a+a \times a r+a r \times \frac{a}{r}=\frac{175}{2}

\Rightarrow \frac{a^{2}}{r}+a^{2} r+a^{2}=\frac{175}{2}

Substituting the value of a

\Rightarrow \frac{25}{r}+25 r+25=\frac{175}{2}

\Rightarrow 50 r^{2}+50 r+50=175 r

\Rightarrow 50 r^{2}-125 r+50=0

\Rightarrow 25\left(2 r^{2}-5 r+2\right)=0

\Rightarrow 2 r^{2}-4 r-r+2=0

\Rightarrow 2 r(r-2)-1(r-2)=0

\Rightarrow(2 r-1)(r-2)=0

\therefore r=\frac{1}{2}, 2

Hence, the G.P. for a=5 and r=\frac{1}{2} is 10,5 and \frac{5}{2}

And, the G.P. for a=5 and r=2 is \frac{5}{2}, 5 and 10

 

QUESTION 5

The sum of first three terms of a G.P. is \frac{39}{10} and their product is 1  . Find the common ratio and the terms.

Sol :

Let the terms of the G.P be \frac{a}{r}, a and ar.

Product of the G.P. = 1

\Rightarrow a^{3}=1

a = 1

Now, sum of the G.P. =\frac{39}{10}

\Rightarrow \frac{a}{r}+a+a r=\frac{39}{10}

\Rightarrow a\left(\frac{1}{r}+1+r\right)=\frac{39}{10}

\Rightarrow 1\left(\frac{1}{r}+1+r\right)=\frac{39}{10}

\Rightarrow 10 r^{2}+10 r+10=39 r

\Rightarrow 10 r^{2}-29 r+10=0

\Rightarrow 10 r^{2}-25 r-4 r+10=0

\Rightarrow 5 r(2 r-5)-2(2 r-5)=0

\Rightarrow(5 r-2)(2 r-5)=0

\Rightarrow r=\frac{2}{5}, \frac{5}{2}

Hence, putting the values of a and r, the required numbers are \frac{5}{2}, 1, \frac{2}{5} or \frac{2}{5}, 1 and \frac{5}{2}

 

QUESTION 6

The sum of three numbers in G.P. is 14  . If the first two terms are each increased by 1 and the third term decreased by 1 , the resulting numbers are in A.P. Find the numbers.

Sol :

Let the numbers be a, ar and ar^2

Sum = 14

\Rightarrow a+a r+a r^{2}=14

\Rightarrow a\left(1+r+r^{2}\right)=14 \quad \ldots(i)

According to the question, a+1,a r+1 and a r^{2}-1 are in A.P.

2(a r+1)=a+1+a r^{2}-1

\Rightarrow 2 a r+2=a+a r^{2}

\Rightarrow 2 a r+2=14-a r [from (i)]

\Rightarrow 3 a r=12

\Rightarrow a=\frac{4}{r}\dots(ii)/

Putting a=\frac{4}{r} in (i)

\Rightarrow \frac{4}{r}\left(1+r+r^{2}\right)=14

\Rightarrow 4 r^{2}-10 r+4=0

\Rightarrow 4 r^{2}-8 r-2 r+4=0

\Rightarrow(4 r-2)(r-2)=0

\Rightarrow r=\frac{1}{2}, 2

Putting r=\frac{1}{2} in (ii), we get a=8

So, the G.P. is 8,4 and 2

Similarly putting r=2 in (ii), we get a=2 .

So, the G.P is 2,4 and 8

 

QUESTION 7

The product of three numbers in G.P. is 216 . If 2,8,6 be added to them, the results are in A.P. Find the numbers.

Sol :

Let the terms of the given G.P. be \frac{a}{r}, a and ar.

Product = 216

\Rightarrow a^{3}=216

a = 6

It is given that \frac{a}{r}+2, a+8 and a r+6 are in A.P.

2(a+8)=\frac{a}{r}+2+a r+6

Putting a=6, we get

\Rightarrow 28=\frac{6}{r}+2+6 r+6

\Rightarrow 28 r=6 r^{2}+8 r+6

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

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

\Rightarrow r=\frac{1}{3}, 3

Hence, putting the values of a and r, the required numbers are 18,6,2 or 2,6 and 18

 

QUESTION 8

Find three numbers in G.P. whose product is 729 and the sum of their products in pairs is 819 

Sol :

Let the required numbers be \frac{a}{r}, a and ar.

Product of the G.P. =729

\Rightarrow a^{3}=729

a = 9

Sum of the products in pairs =819

\Rightarrow \frac{a}{r} \times a+a \times a r+a r \times \frac{a}{r}=819

\Rightarrow a^{2}\left(\frac{1}{r}+r+1\right)=819

\Rightarrow 81\left(\frac{1+r^{2}+r}{r}\right)=819

\Rightarrow 9\left(r^{2}+r+1\right)=91 r

\Rightarrow 9 r^{2}-82 r+9=0

\Rightarrow 9 r^{2}-81 r-r+9=0

\Rightarrow(9 r-1)(r-9)=0

\Rightarrow r=\frac{1}{9}, 9

Hence, putting the values of a and r, we get the numbers to be 81,9 and 1 or 1,9 and 81

 

QUESTION 9

The sum of three numbers in G.P. is 21 and the sum of their squares is 189  Find the numbers.

Sol :

Let the required numbers be a, ar and a r^{2}

Sum of the numbers =21

\Rightarrow a+a r+a r^{2}=21

\Rightarrow a\left(1+r+r^{2}\right)=21 \quad \ldots(\mathrm{i})

Sum of the squares of the numbers =189

\Rightarrow a^{2}+(a r)^{2}+\left(a r^{2}\right)^{2}=189

\Rightarrow a^{2}+(a r)^{2}+\left(a r^{2}\right)^{2}=189

\Rightarrow a^{2}\left(1+r^{2}+r^{4}\right)=189 \quad \ldots( ii )

Now, a\left(1+r+r^{2}\right)=21 [from(i)]

Squaring both the sides

\Rightarrow a^{2}\left(1+r+r^{2}\right)^{2}=441

\Rightarrow a^{2}\left(1+r^{2}+r^{4}\right)+2 a^{2} r\left(1+r+r^{2}\right)=441

\Rightarrow 189+2 a r\left\{a\left(1+r+r^{2}\right)\right\}=441 [using (ii)]

\Rightarrow 189+2 a r \times 21=441 [using (i)]

\Rightarrow a r=6

\Rightarrow a=\frac{6}{\mathrm{r}}

Putting a=\frac{6}{r} in (\mathrm{i})

\frac{6}{r}\left(1+r+r^{2}\right)=21

\Rightarrow \frac{6}{r}+6+6 r=21

\Rightarrow 6 r^{2}+6 r+6=21 r

\Rightarrow 6 r^{2}-15 r+6=0

\Rightarrow 3\left(2 r^{2}-5 r+2\right)=0

\Rightarrow 2 r^{2}-5 r+2=0

\Rightarrow(2 r-1)(r-2)=0

\Rightarrow r=\frac{1}{2}, 2

Putting r=\frac{1}{2} in a=\frac{6}{r}, we get a=12

So, the numbers are 12,6 and 3

Putting r=2 in a=\frac{6}{r}, we get a=3

So, the numbers are 3,6 and 12 .

Hence, the numbers that are in G.P are 3,6 and 12 .

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