This set of Discrete Mathematics Multiple Choice Questions & Answers focuses on “Arithmetic and Geometric Mean”.

1. Let A_{1}, A_{2}, be two AM’s and G_{1}, G_{2} be two GM’s between a and b,then (A_{1} + A_{2}) / G_{1}G_{2} is equal to :

a) (a+b) / 2ab

b) 2ab/(a+b)

c) (a+b)/(ab)

d) None of the mentioned

View Answer

Explanation: A

_{1}+ A

_{2}= a + b, G

_{1}G

_{2}= ab.

2. The series a,(a+b)/2, b is in

a) AP

b) GP

c) HP

d) None of the mentioned

View Answer

Explanation: (a+b)/2 is AM between a, b. Hence series is in AP.

3. The series a, (ab)^{1/2}, b is in

a) AP

b) GP

c) HP

d) None of the mentioned

View Answer

Explanation: (ab)

^{1/2}is GM between a, b. Hence series is in GP.

4. If A and G be the A.M and G.M between two positive number then the numbers are A + (A^{2} – G^{2})^{1/2}, A – (A^{2} – G^{2})^{1/2} . The given statement is

a) True

b) False

View Answer

Explanation: The equation having its roots as given equation is

x

^{2}– 2Ax + G

^{2}= 0 which implies

x = A + (A

^{2}– G

^{2})

^{1/2}, A – (A

^{2}– G

^{2})

^{1/2}.

5. If one geometric mean G and two airthmetic mean A_{1}, A_{2} are inserted between two numbers,then (2A_{1} – A_{2}) (2A_{2} – A_{1}) is equal to

a) 2G

b) G

c) G^{2}

4) None of the mentioned.

View Answer

Explanation: Let a and b be two numbers then, G = (ab)

^{1/2}, A

_{1}= (2a+b)/3, A

_{2}= (a+2b)/3, (2A

_{1}– A

_{2}) = a, (2A

_{2}– A

_{1}) = b, (2A

_{1}– A

_{2})(2A

_{2}– A

_{1}) = G

^{2}.

6. State whether the given statement is true or false

AM ≤ GM.

a) True

b) False

View Answer

Explanation: Airthmetic Mean is always greater or equal to geometric mean.

7. If between two numbers which are root of given equation

x^{2} – 18x + 16 = 0, a GM is inserted then the value of that GM is.

a) 4

b) 5

c) 6

d) 16

View Answer

Explanation: x

^{2}– 2Ax + G

^{2}= 0, here G

^{2}= 16 and therefore G = 4.

8. If a_{1}, a_{2}, a_{3} are in airthemetic as well as geometric progression then which of the following is/are correct ?

a) 2a_{2} = a_{1} + a_{3}

b) a_{2} = (a_{1}a_{3})^{1/2}

c) a_{2} – a_{1} = a_{3} -a_{2}

d) All of the mentioned are correct.

View Answer

Explanation: a

_{2}is AM, GM between a

_{1}, a

_{3}, also the series is in AP so common difference should be same.

9. If a_{1}, a_{2}, a_{3} are in GP then 1/a_{1}, 1/a_{2}, 1/a_{3} are in :

a) AP

b) GP

c) HP

d) None of the mentioned.

View Answer

Explanation: Let the terms be ar, a, a/r then reciprocals are 1/(ar), 1/a, r/a. Still the terms are in GP.

10. If a_{1}, a_{2}, a_{3}…….. are in AP then if a_{7} = 15,then the value of common difference that would make a_{2} a_{7} a_{12} greatest is

a) 2

b) 0

c) 4

d) 9

View Answer

Explanation: Let d be common difference of the AP. Then,

a

_{2}a

_{7}a

_{12}= (15 – 5d)(15)(15 + 5d) = 375(9 – d

^{2})

For maximum value d=0.

**Sanfoundry Global Education & Learning Series – Discrete Mathematics.**

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