Test: Relation Between AM And GM - Commerce MCQ

# Test: Relation Between AM And GM - Commerce MCQ

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## 10 Questions MCQ Test Mathematics (Maths) Class 11 - Test: Relation Between AM And GM

Test: Relation Between AM And GM for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: Relation Between AM And GM questions and answers have been prepared according to the Commerce exam syllabus.The Test: Relation Between AM And GM MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Relation Between AM And GM below.
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Test: Relation Between AM And GM - Question 1

### The G.M. between the numbers: 56 and 14 is:

Detailed Solution for Test: Relation Between AM And GM - Question 1

Geometric mean of two numbers a and b is √ab
As two numbers are 14 and 56
Geometric mean is √14×56
= ± √2×7×2×2×2×7
= ±(2×2×7)
= ±28

Test: Relation Between AM And GM - Question 2

### Three geometric means between the numbers 1/4 and 64 are:

Detailed Solution for Test: Relation Between AM And GM - Question 2

nth G.M. between a and b is
Gn = arn
Where common ratio is r = (b/a)(1/(n+1))
​So, to insert 3 geometric means between 1/4 and 64
r = (b/a)(1/(n+1))
r = (64/(¼)(1/(3+1))
r = (256)1/4
r  = (4)(4)1/4
r = 4
Gn = 1 * (4)n
G0 = 1 * (4)0 = 1
G1 = 1 * (4)1 = 4
G2 = 1 * (4)2 = 16
The terms are 1, 4, 16

Test: Relation Between AM And GM - Question 3

### The three numbers between 1 and 256 such that the sequence is in GP are

Detailed Solution for Test: Relation Between AM And GM - Question 3

nth G.M. between a and b is
Gn = arn
Where common ratio is r = (b/a)(1/(n+1))
​So, to insert 3 geometric means between 1 and 256
r = (b/a)(1/(n+1))
r = (256/(1)(1/(3+1))
r = (256)1/4
r  = (4)(4)1/4
r = 4
Gn = 1 * (4)n
G1 = 1 * (4)1 = 4
G2 = 1 * (4)2 = 16
G3 = 1 * (4)3 = 64
The terms are  4, 16, 64

Test: Relation Between AM And GM - Question 4

If A and G are A.M. and G.M. of two real numbers a and b, then

Detailed Solution for Test: Relation Between AM And GM - Question 4

A=a+b/2. ,G=√ab
A-G=(a+b/2)-√ab
=(a+b-2√ab)/2
=(√a-√b)2/2 is greater than or equal to zero
A-G is greater than or equal to zero so
So A is greater than equal to zero

Test: Relation Between AM And GM - Question 5

The sum of the series 2 + 6 + 18 + ….+ 4374 is:

Detailed Solution for Test: Relation Between AM And GM - Question 5

The given series is a geometric series in which a=2,r=3,l=4374.
Therefore,
Required sum = (lr−a)/(r−1)
​= (4374×3−2)/(3−1)
​= 6560

Test: Relation Between AM And GM - Question 6

If a, b, c are in A.P. and k is any non zero numbre, then ka, kb, kc are in

Detailed Solution for Test: Relation Between AM And GM - Question 6

a, b, c are in AP
Let d be the common difference.
b = a + d,
c = a + 2d

There are in GP with common ratio: kd

Test: Relation Between AM And GM - Question 7

The A.M. between two numbers is 34 and their G.M. is 16.The numbers are

Detailed Solution for Test: Relation Between AM And GM - Question 7

Let the numbers be x, y

Then arithmatic mean = (x+y)/2 =34

→x+y =68

Also geometric mean =√(xy)=16

oy xy=16^2=256

Hence

x(68−x)=256

or x^2−68x+256=0

(x−64)(x−4)=0

Hence x=64 or x=4

and y=4 or 64

Larger number is 64

Test: Relation Between AM And GM - Question 8

Two geometric means g and g’ and one arithmetic mean A is inserted between two numbers, then

Detailed Solution for Test: Relation Between AM And GM - Question 8

Test: Relation Between AM And GM - Question 9

The G.M. between 3/2 and 27/2 is

Detailed Solution for Test: Relation Between AM And GM - Question 9

b2 = ac
a = 3/2, c = 27/2
⇒ b2 = 3/2 * 27/2
⇒ b = (√81/4)
⇒ b = 9/2

Test: Relation Between AM And GM - Question 10

How many terms of geometric progression 4 , 16 , 64 , are required to give the sum 5460?

Detailed Solution for Test: Relation Between AM And GM - Question 10

a(rn-1)/(r-1) = 5460
=> 4(4n-1)/4-1 = 5460
=> 4n = 4096
=> 22n = 212
=> n = 6

## Mathematics (Maths) Class 11

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## Mathematics (Maths) Class 11

75 videos|238 docs|91 tests