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3, a, b, 192    are terms in G.P? Find a and b.

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seedrasoso | Student, Grade 10 | eNoter

Posted August 24, 2010 at 10:15 AM via web

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3, a, b, 192    are terms in G.P? Find a and b.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted August 24, 2010 at 10:18 AM (Answer #1)

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3, a, b, 192   are parts of a geometric progression:

a1= 3

a2= 3*r = a

a3= 3*r^2 = b

a4= 3*r^3 = 192

==> 3r^3 = 192

Divide by 3:

==> r^3 = 192 /3 = 64

==> r^3 = 64

Now take the cubic root for both sides:

==> r= 4

Now let us calculater a and b

a2= a = 3*r = 3*4 = 12

a3= b = 3*r^2 = 3*4^2 = 3*16 = 48

==> a= 12    and   b = 48

3, 12, 48, 192 are terms in a G.P where r = 4

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krishna-agrawala | College Teacher | Valedictorian

Posted August 24, 2010 at 10:40 AM (Answer #2)

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nth term of a geometric progression (G.P.) takes the general form:

nth term = tn= p*q^(n -1)

Where p and q are constants.

Using this general form the 1st terms of the given series can be written as:

t1 = 3 = p*q^(1 - 1) = p*q^0 = p*1 = p

Therefore:

p = 3

Using this value of p in the general form of G.P. the other three terms of given series can be expressed as:

t2 = a = 3*q^(2 - 1) = 3*q^1 = 3*q

t3 = b = 3*q^(3 - 1) = 3*q^2

t3 = 192 = 3*q^(4 - 1) = 3*q^3

==> 192 = 3*q^3

==> q^3 = 192/3 = 64

Therefore:

q = 64^1/3 = 4

Therefore:

a = 3*4 = 12

b = 3*4^2 = 3*16 = 48

Answer:

a = 12

b = 48

As all the four terms can be expressed in general form for a G.P., these are in G.P.

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neela | High School Teacher | Valedictorian

Posted August 24, 2010 at 11:29 AM (Answer #3)

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Since 3,a,b and 192 are in GP, the consecutive terms has the same common ratio r.

Therefore

3/a =  a/b = b/192 = r.

Therefore

3 = ar,

a = br

b = 192r.

3 = br*r = 192r*r*r = 192r^3.

r^3 = 3/192 = 1/64.

r = cube root (1/64) = 1/4.

Therefore a = 3/r = 3/(1/4) = 12.

b = a/r = 12/(1/4) = 48.

So a = 12 and b = 48.

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giorgiana1976 | College Teacher | Valedictorian

Posted August 24, 2010 at 5:18 PM (Answer #4)

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Because 3, a, b, 192 are the terms of a geometric sequence, we'll use the theorem of geometric mean:

a^2 = 3*b (1)

b^2 = a*192 (2)

b = sqrt (192a) (3)

We'll substitute b in (1)

a^2 = 3*sqrt (192a)

Now, we'll write each term of g.p.:

a = 3r, where r is the common ratio

b = a*r = 3r*r = 3r^2

192 = b*r

192 = 3r^2 *r

192 = 3r^3

We'll divide by 3:

64 = r^3

r = (64)^1/3

r = 4

So, a = 3*r = 3*4 = 12

a = 12

We'll substitute a in (3):

b = sqrt (192a)

b  = sqrt (192*12)

b = sqrt 2304

b = 48

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manager25 | Student, Grade 9 | eNoter

Posted August 25, 2010 at 2:13 AM (Answer #5)

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3, a, b, 192   are parts of a gp

a1= 3 (1st term)

a2= 3*r = a (2nd term), where are is the ratio

a3= 3*r^2 = b (3rd term)

a4= 3*r^3 = 192 (4th term)

==> 3r^3 = 192

Divide by 3:

==> r^3 = 192 /3 = 64

==> r^3 = 64

Now take the cubic root for both sides:

==> r= 4

Now let us calculater a and b

a2= a = 3*r = 3*4 = 12

a3= b = 3*r^2 = 3*4^2 = 3*16 = 48

==> a= 12    and   b = 48

3, 12, 48, 192 are terms in a G.P where r = 4

 

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