Description

This mock test of Test: Geometric Progressions for JEE helps you for every JEE entrance exam.
This contains 10 Multiple Choice Questions for JEE Test: Geometric Progressions (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Geometric Progressions quiz give you a good mix of easy questions and tough questions. JEE
students definitely take this Test: Geometric Progressions exercise for a better result in the exam. You can find other Test: Geometric Progressions extra questions,
long questions & short questions for JEE on EduRev as well by searching above.

QUESTION: 1

For a G.P. the ratio of the 7^{th} and the third terms is 16. The sum of 9 terms is 2555. What is the first term?

Solution:

ar^{6}/ar^{2} = 16

r^{4} = 16

r^{4} = (2)^{4}

r = 2

S_{9} = a(r^{n} - 1)

2555 = a((2)^{9} - 1)

2555 = a(512-1)

2555/511 = a

a = 5

QUESTION: 2

The G.M. of 5 and 8 is

Solution:

The geometric mean of two numbers, say x, and y is the square root of their product x * y.

Mean = [x * y]^{1/2}

= [5 * 8]^{1/2}

= [40]^{1/2}

= 2(10)^{1/2}

QUESTION: 3

The 10^{th} term of an G.P. is and the 5th term is.What is the Common Ratio?

Solution:

ar^{9 }= 1/(2)^{8} -----------------------(1)

ar^{3} = 1/(2)^{2} ------------------------------(2)

Comparing eq (1) & (2)

1/(2)^{8} r^{9} = 1/(2)^2 r^{4}

1/r^{5} = (2)^{5}

r^{5} = 1/(2)^{5}

r = 1/2

QUESTION: 4

A sequence a_{1}, a_{2}, a_{3},…, a_{n} is called ______ progression, if each term is non-zero and = constant for

Solution:

The expression depicts the ratio of two numbers, if the ratio between the numbers is constant, then it will definately form a GP.

QUESTION: 5

Which of the following sequeces in GP will have common ratio 3,where n is an Integer?

Solution:

g_{n} = 6( 3^{n-1}) it is a geometric expression with coefficient of constant as 3^{n-1}.So it is GP with common ratio 3.

QUESTION: 6

The third term of a G.P. is 3, the product of first five terms of this progression is:

Solution:

QUESTION: 7

The sum of n terms of the sequence 8, 88, 888,…. is:

Solution:

8(1) + 8(11) + 8(111) + 8(1111) +........up to n times

= 8{ 1 + 11 + 111 + 1111 + ....... up to n times}

= 8/9{9 + 99 + 999 + 9999 + ..... up to n times}

= 8/9 {(10-1 )+ (100-1) + (1000-1) + (10000-1) + ........up to n times}

= 8/9 {(10 + 100 + 1000 + 10000 +...... n times) - (n×1)}

= 8/9{( (10×(10^{n} -1))/(10-1)) -n}

= 80/81(10^{n} - 1) - n

= 8/81(10^{n+1} - 10 -9n)

QUESTION: 8

Which term of the following sequence is 64 ?

2 , 2√2, 4, .....

Solution:

Given sequence : 2, 2√2, 4….

First term a_{1} = a = 2 and 2nd term a_{2} = 2√2, then

Common ratio r = a_{2}/a = (2√2)/2

Let a_{n} = 64

∴ ar^{(n-1)} = 64

⇒ 2.(√2)^{(n-1)} = 32

⇒ (2)^{(n-1)/2 }= 32

∴ (2)^{(n-1)/2} = (2)^{5}

⇒ (n − 1)/2 = 5,

⇒ n = 11

QUESTION: 9

How many terms of the G.P. 4 + 16 + 64 + … will make the sum 5460?

Solution:

Sum (S_{n}) = a x (r^{n} -1)/(r-1)

5460 = 4 x (4^{n} -1)/3

16380 = 4^{n+1} - 4

16384 = 4^{n+1}

4^{7 }= 4^{n+1}

7 = n + 1

n = 6

QUESTION: 10

What is the 50^{th} term of the sequence √3, 3, 3√3, 9, ......

Solution:

an = ar^{(n-1)}

Given, a = √3, r = 3/√3

r = √3

a_{50} = ar^{(n-1)}

= (√3)(√3)^{(50-1)}

= (√3)(√3)^{49}

= (√3)^{50}

### Geometric Modeling

Doc | 4 Pages

### Arithmetic progressions

Doc | 4 Pages

### Geometric Sequences

Video | 01:37 min

### Progressions Formula

Doc | 1 Page

- Test: Geometric Progressions
Test | 10 questions | 10 min

- Test: Arithmetic And Geometric Progressions - 1
Test | 40 questions | 40 min

- Test: Arithmetic And Geometric Progressions - 3
Test | 40 questions | 40 min

- Test: Arithmetic And Geometric Progressions - 4
Test | 40 questions | 40 min

- Test: Arithmetic And Geometric Progressions - 2
Test | 40 questions | 40 min