The document Relationship Between A.M. & G.M. CA Foundation Notes | EduRev is a part of the CA Foundation Course Quantitative Aptitude for CA CPT.

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Let a and b be the two numbers.

Let A and G be the A. M. and G. M. respectively between a and b

âˆ´ A > G

**Example 1. The arithemetic mean between two numbers is 34 and their geometric meanis 16. Find the numbers.**

**Solution : **Let the numbers be a and b.

Since A. M. between a and b is 34,

.....(1)

Since G. M. between a and b is 16,

âˆ´ âˆšab = 16 or, ab = 256

we know that (a â€“ b)^{2} = (a + b)^{2} â€“ 4 ab ......(2)

= (68)^{2} â€“ 4 Ã— 256

= 4624 â€“ 1024 = 3600

âˆ´ a â€“ b = âˆš3600 = 60 ....... (3)

Adding (1) and (3), we get, 2a = 128

âˆ´ a = 64

Subtracting (3) from (1), are get

2b = 8 or, b = 4

âˆ´ Required numbers are 64 and 4.

**Example 2. The arithmetic mean between two quantities b and c is a and the twogeometric means between them are g _{1} and g_{2}. Prove that**

**Solutions: **The A. M. between b and c is a

Again g_{1} and g_{2} are two G. M.'s between b and c

âˆ´ b, g_{1}, g_{2}, c are in G. P.

If r be the common ratio, then

c = br^{3}

or

= bc (2a) [since b + c = 2a]

= 2abc

**Example 3. If one geometric mean G and two arithmetic means p and q be inserted between two quantities, show that : G ^{2} = (2p â€“ q) (2q â€“ p) **

**Solutions: **Let the two quantities be a and b, then

G = âˆšab or, G^{2} = ab ...... (1)

Also p and q are two A. M.'s betweena and b

âˆ´ a, p, q, b are in A. P.

âˆ´ p â€“ a = q â€“ p and q â€“ p = b â€“ q

âˆ´ a = 2p â€“ q and b = 2q â€“ p

âˆ´ G^{2} = ab = (2p â€“ q) (2q â€“ p)

**Example 4. The product of first three terms of a G. P. is 1000. If we add 6 to its secondterm and 7 to its 3rd term, the three terms form an A. P. Find the terms of the G. P.**

**Solutions: **Let t_{1} = a/r, t_{2} = a and t_{3} = ar be the first three terms of G. P.

Then, their product = a/r. a.ar = 1000 or a^{3} = 1000 or a = 10

By the question, t_{1}, t_{2} + 6, t_{3} + 7 are in A. P. ...... (1)

i.e, a/r, a + 6, ar + 7 are in A.P.

âˆ´ (a + 6) â€“ a/r = (ar + 7) â€“ (a + 6)

or (a + 6) = a/r + (ar + 7)

or, 2(10+6) = 10/r + (10r + 7)

or, 32r = 10 + 10 r^{2} + 7r

or, 10r^{2} â€“ 25r + 10 = 0

= 2, 1/2

When a = 10, r = 2. then the terms are 10/2, 10(2) i.e. 5,10,20

When a = 10, r = 1/2, then the terms are 10(2), 10, 10(1/2)

i.e., 20, 10, 5

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