Geometric Progression Questions and Answers

First term a = 1 and r = 9.

s_{n} = a \times \frac{r^{n} -1}{r-1} if r>1

s_{7} = 1 \times \frac{9^{7} -1}{9-1}

= 5,97,871

For this series, we have a = 3, r = 3 and n = 5.

s_{n} = a \times \frac{r^{n} -1}{r-1} if r>1

s_{5} = 3 \times \frac{3^{5} -1}{3-1} = 363

For this series, we have a = 8, r = \frac{-1}{2} and n = 7.

Thus

s_{n} = a \times \frac{1-r^{n}}{1-r} if r<1

s_{7} = 8 \times \frac{1-(\frac{-1}{2})^{7}}{1-(\frac{-1}{2})}

S₇ =5.37.

Here a = 4 and r = 2. nth term=512. But the formula for the nth term is ar^{n-1}

So

512 = 4\times 2^{n-1}

128= 2^{n-1}

2^{7} = 2^{n-1}

7 = n − 1

n = 8.

The series of gp is: 1, ar², ar³,………

Common ratio=3

Last term= 486

So a_{n} = a\times r^{n-1} = 486

a_{n} = a\times 3^{n-1}=486

a(3ⁿ) = 486\times 3 = 1458-----(1)

Now, sum of G.P.=

=  728

By putting the value of r and using eq(1) we get

\frac{1458-a}{2} = 728

1458 - a = 1456

So a= 2.

As we know first term a=1 and r=5.

s_{n} = a \times \frac{r^{n} -1}{r-1} if r>1

s_{10} = 1 \times \frac{5^{10} -1}{5-1}

= 24,41,406

As we know first term a=5 and r=3.

Formula= s_{n} = a \times \frac{r^{n} -1}{r-1} if r>1

s_{17} = 5 \times \frac{3^{17} -1}{3-1}

=32,28,50,405

Common ratio = 6 and  First term = 6

As per the formula:

s_{n} = a \times \frac{r^{n} -1}{r-1} if r>1

s_{6} = 6 \times \frac{6^{6} -1}{6-1}

= 55986

Here a=3

r=0.75

Since |r|<1

So = \frac{a_{1}}{1-r}

Or, sum= \frac{3}{(1-0.75)} = 12

a_{n} = a\times r^{n-1}

here, r = \frac{1}{2} and n=4

so the series will be:

a_{1},\frac{a_{1}}{2},\frac{a_{1}}{4},\frac{a_{1}}{8}

For this series, we have a = 3, r = 3 and n = 5.

s_{n} = a \times \frac{r^{n} -1}{r-1} if r>1

s_{5} = 3 \times \frac{3^{5} -1}{3-1} = 363

For this series, we have a = 8, r = \frac{-1}{2} and n = 7.

Thus

s_{n} = a \times \frac{1-r^{n}}{1-r} if r<1

s_{7} = 8 \times \frac{1-(\frac{-1}{2})^{7}}{1-(\frac{-1}{2})}

S₇ =5.37.

Here a = 4 and r = 2. nth term=512. But the formula for the nth term is ar^{n-1}

So

512 = 4\times 2^{n-1}

128= 2^{n-1}

2^{7} = 2^{n-1}

7 = n − 1

n = 8.

The series of gp is: 1, ar², ar³,………

Common ratio=3

Last term= 486

So a_{n} = a\times r^{n-1} = 486

a_{n} = a\times 3^{n-1}=486

a(3ⁿ) = 486\times 3 = 1458-----(1)

Now, sum of G.P.=

=  728

By putting the value of r and using eq(1) we get

\frac{1458-a}{2} = 728

1458 - a = 1456

So a= 2.

As we know first term a=1 and r=5.

s_{n} = a \times \frac{r^{n} -1}{r-1} if r>1

s_{10} = 1 \times \frac{5^{10} -1}{5-1}

= 24,41,406

As we know first term a=5 and r=3.

Formula= s_{n} = a \times \frac{r^{n} -1}{r-1} if r>1

s_{17} = 5 \times \frac{3^{17} -1}{3-1}

=32,28,50,405

Common ratio = 6 and  First term = 6

As per the formula:

s_{n} = a \times \frac{r^{n} -1}{r-1} if r>1

s_{6} = 6 \times \frac{6^{6} -1}{6-1}

= 55986

Here a=3

r=0.75

Since |r|<1

So = \frac{a_{1}}{1-r}

Or, sum= \frac{3}{(1-0.75)} = 12

First term a_{1} = 20 , common ratio, r = \frac{1}{5} = 0.2

Number of terms , n=10

Sum of first 10 terms is

s_{n} = a \times \frac{1-r^{n}}{1-r} if r<1 [latex]s_{10} = 20 \times \frac{1-0.2^{10}}{1-0.2} [/latex] Sum of first 10 terms is 24.99.

(abc)^{\frac{1}{3}}

Here series will be like: a, ar, ar^{2} , ar^{3}

or, ar+ar²=12 and ar²+ar³=60

or,\frac{ar^{2}+ar^{3}}{ar+ar^{2}} = 5

or r+r² = 5(1+r)

r²+r = 5r+5

r²-4r-5 = 0

(r+1)(r-5) = 0

So r = either 5 or -1

As we know the formula of nth term of GP = ar^n-1
9th term = ar⁸ and 5th term =ar⁴
Given that
ar⁸ = 16ar⁴
=> r⁴ = 16
=> r = 2
Given ar⁶ = 96
put value of r above
a(2)⁶ = 96
a = \frac{96}{64} = 1.5
Hence, first term of the GP is 1.5.

As we know: in geometric mean problems ab = \sqrt{ab}

Here a = 30, b = 45

G.M. of ab = \sqrt{ab} = 36.74

The nth term of a geometric sequence is a_{n} = a\times r^{n-1}

Here a is the first term and r the common ratio

r = \frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} =......=\frac{a_{n}}{a_{n-1}}

Now a=4 and r = \frac{12}{4} = 3

So = a_{9} = 4\times 3^{8} = 26244

The ratio of each term with earlier term is same.

So , Geometric Progression

In g(n) = 6(3^{n-1}) Geometric Progression series as it consists coefficient of 3^{n-1}

Let \frac{x}{r},  x , xr be three terms ,then \frac{x}{r} \times x \times xr = 8

x = 2

The term y should be the Geometric Mean of term x and z.