# In a set of four numbers the first three are in G.P, and the last three in A.P, with common difference 6. If the first number is the same as the fourth, find the four numbers.

Last updated date: 18th Mar 2023

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Answer

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Hint: - If$a,b,c.....................{\text{G}}{\text{.P}}$, then use property ${b^2} = ac$

Let the four numbers be a, b, c, d.

It is given that last three numbers are in A.P with common difference 6

$ \Rightarrow b,{\text{ }}c,{\text{ d}}..........................{\text{A}}{\text{.P}}$

Now we know in A.P the difference of second number to first number is equal to difference of third number to second number is equal to common difference which is 6 (given).

$

\Rightarrow c - b = d - c = 6 \\

\Rightarrow c = b + 6,{\text{ }}d = c + 6 \\

\Rightarrow c = b + 6,{\text{ }}d = b + 6 + 6 = b + 12 \\

$

Now, it is given that first number is same as fourth number

$ \Rightarrow a = d = b + 12$

So, the numbers is

$b + 12,{\text{ }}b,{\text{ }}b + 6{\text{, }}b + 12$

It is given that first three numbers are in G.P

\[b + 12,{\text{ }}b,{\text{ }}b + 6........................G.P\]

Now according to G.P property second number square is equal to first times third number

\[

\Rightarrow {b^2} = \left( {b + 12} \right)\left( {b + 6} \right) \\

\Rightarrow {b^2} = {b^2} + 6b + 12b + 72 \\

\Rightarrow 18b = - 72 \\

\Rightarrow b = - 4 \\

\]

So the numbers are $ - 4 + 12,{\text{ - 4}},{\text{ - 4}} + 6{\text{, - 4}} + 12$

$ = 8,{\text{ }} - 4,{\text{ }}2,{\text{ 8}}$

So, these are the required four numbers.

Note: - In such types of questions the key concept we have to remember is that always remember the properties of Arithmetic Progression and Geometric Progression which is stated above, so after applying these properties and according to given conditions we will get the required answer.

Let the four numbers be a, b, c, d.

It is given that last three numbers are in A.P with common difference 6

$ \Rightarrow b,{\text{ }}c,{\text{ d}}..........................{\text{A}}{\text{.P}}$

Now we know in A.P the difference of second number to first number is equal to difference of third number to second number is equal to common difference which is 6 (given).

$

\Rightarrow c - b = d - c = 6 \\

\Rightarrow c = b + 6,{\text{ }}d = c + 6 \\

\Rightarrow c = b + 6,{\text{ }}d = b + 6 + 6 = b + 12 \\

$

Now, it is given that first number is same as fourth number

$ \Rightarrow a = d = b + 12$

So, the numbers is

$b + 12,{\text{ }}b,{\text{ }}b + 6{\text{, }}b + 12$

It is given that first three numbers are in G.P

\[b + 12,{\text{ }}b,{\text{ }}b + 6........................G.P\]

Now according to G.P property second number square is equal to first times third number

\[

\Rightarrow {b^2} = \left( {b + 12} \right)\left( {b + 6} \right) \\

\Rightarrow {b^2} = {b^2} + 6b + 12b + 72 \\

\Rightarrow 18b = - 72 \\

\Rightarrow b = - 4 \\

\]

So the numbers are $ - 4 + 12,{\text{ - 4}},{\text{ - 4}} + 6{\text{, - 4}} + 12$

$ = 8,{\text{ }} - 4,{\text{ }}2,{\text{ 8}}$

So, these are the required four numbers.

Note: - In such types of questions the key concept we have to remember is that always remember the properties of Arithmetic Progression and Geometric Progression which is stated above, so after applying these properties and according to given conditions we will get the required answer.

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