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# Test: Geometric Progressions

## 10 Questions MCQ Test Mathematics For JEE | Test: Geometric Progressions

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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 7th and the third terms is 16. The sum of 9 terms is 2555. What is the first term?

Solution:

ar6/ar2 = 16
r4 = 16
r4 = (2)4
r = 2
S9 = a(rn - 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
= 1/2
= 2(10)1/2

QUESTION: 3

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

Solution:

ar9 = 1/(2)8 -----------------------(1)
ar3 = 1/(2)2 ------------------------------(2)
Comparing eq (1) & (2)
1/(2)8 r9 = 1/(2)^2 r4
1/r5 = (2)5
r5 = 1/(2)5
r = 1/2

QUESTION: 4

A sequence a1, a2, a3,…, an 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:

gn = 6( 3n-1) it is a geometric expression with coefficient of constant as 3n-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×(10n -1))/(10-1)) -n}
= 80/81(10n - 1) - n
=  8/81(10n+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 a1 = a = 2 and 2nd term a2 = 2√2, then
Common ratio r = a2/a = (2√2)/2
Let an = 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 (Sn) = a x (rn -1)/(r-1)
5460 = 4 x (4n -1)/3
16380 = 4n+1 - 4

16384 = 4n+1

4= 4n+1

7 = n + 1

n = 6

QUESTION: 10

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

Solution:

an = ar(n-1)
Given, a = √3, r = 3/√3
​r = √3
a50 = ar(n-1)
= (√3)(√3)(50-1)
= (√3)(√3)49
= (√3)50