Question

Q.2) Solve the following question.
If 4a2b.Bab2 ,p are in continued proportion then find the value of p.​

Answer 1

Answer:

SOLUTION :

GIVEN

\sf{ 4 {a}^{2}b ,8a {b}^{2} ,p\: \: are \: in \: continued \: proportion}4a

2

b,8ab

2

,pareincontinuedproportion

TO DETERMINE

The value of p

CONCEPT TO BE IMPLEMENTED

If three numbers a, b, c are in continued proportion then

a : b = b : c

EVALUATION

It is given that

\sf{ 4 {a}^{2}b ,8a {b}^{2} ,p\: \: are \: in \: continued \: proportion}4a

2

b,8ab

2

,pareincontinuedproportion

Hence

\sf{ 4 {a}^{2}b : 8a {b}^{2} =8a {b}^{2} : p}4a

2

b:8ab

2

=8ab

2

:p

\implies \displaystyle \sf{ \frac{4 {a}^{2}b }{8a {b}^{2}} \: } = \sf{ \frac{8a {b}^{2}}{p} }⟹

8ab

2

4a

2

b

=

p

8ab

2

\implies \displaystyle \sf{ p \times {4 {a}^{2}b } = {8a {b}^{2}} \: } \times \sf{{8a {b}^{2}} }⟹p×4a

2

b=8ab

2

×8ab

2

\implies \displaystyle \sf{ p \times {4 {a}^{2}b } = 64 {a}^{2} {b}^{4} }⟹p×4a

2

b=64a

2

b

4

\implies \displaystyle \sf{ p = 16 {b}^{3} }⟹p=16b

3

RESULT

Hence the required answer is

\boxed{ \displaystyle \sf{ \: \: p = 16 {b}^{3} \: }}

p=16b

3

Step-by-step explanation:

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Answer 2

Answer:

$p = 16b^3$

Step-by-step explanation:

Given that, $4a^2b$, $8ab^2$, $p$ are in continued proportion.

∴ mean squared = product of extremes

∴ $(8ab^2)^2 = 4a^2b. p$

$=> \frac{4a^2b. p}{4a^2b} = \frac{(8ab^2)^2}{4a^2b}$  [swapping sides and dividing both sides by $4a^2b$]

∴ $p = 16b^3$