Three numbers whose sum is 21 are in AP. If 2, 2, 4 are added to them respectively, the resulting numbers are in GP. Find the numbers.
Answer
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Hint: In this problem, we are given that three numbers whose sum is 21 are in AP. If 2, 2, 4 are added to them respectively, the resulting numbers are in GP, we have to find the numbers. Here we can first assume three numbers and add the three numbers which are equal to 21, to get the value of a. We can then add the given three numbers to them. We can then use one of the geometric progression forms to simplify the new three numbers to get the number in GP.
Complete step by step answer:
Here we are given that three numbers whose sum is 21 are in AP. If 2, 2, 14 are added to them respectively, the resulting numbers are in GP, we have to find the numbers.
We can now assume the three number as,
\[a-d,a,a+d\]
We are given that the sum of three number is 21, we can write it as,
\[\begin{align}
& \Rightarrow a=d+a+a-d=21 \\
& \Rightarrow a=7 \\
\end{align}\]
We can now substitute the value of a in the assumed three numbers, we get
\[\Rightarrow 7-d,7,7+d\] …….. (1)
We can now add the given three numbers, 2, 2, 14 in the above numbers, we get
\[\begin{align}
& \Rightarrow {{t}_{1}}=7+2-d=9-d \\
& \Rightarrow {{t}_{2}}=7+2=9 \\
& \Rightarrow {{t}_{3}}=7+14+d=21+d \\
\end{align}\]
We can see that the above terms are in GP.
We know that in GP, we have
\[\dfrac{{{t}_{2}}}{{{t}_{3}}}=\dfrac{{{t}_{1}}}{{{t}_{2}}}\]
We can now substitute the values in the above step, we get
\[\Rightarrow \dfrac{9}{21+d}=\dfrac{9-d}{9}\]
We can now simplify the above step by cross multiplication, we get
\[\begin{align}
& \Rightarrow 81=\left( 21+d \right)\left( 9-d \right) \\
& \Rightarrow {{d}^{2}}+12d-108=0 \\
& \Rightarrow \left( d+18 \right)\left( d-6 \right)=0 \\
\end{align}\]
We can now get the value of d,
\[d=-18,6\]
Therefore, the numbers in AP are
when d = -18, then from (1)
\[\Rightarrow 25,7,-11\]
When d = 6, then from (1), we get
\[\Rightarrow 1,7,13\]
Note: We should always remember that when the numbers are in GP, then we have some identities to solve these types of problems. We should first find the value of a from the given data to solve for d and find the numbers using the geometrical progression identities.
Complete step by step answer:
Here we are given that three numbers whose sum is 21 are in AP. If 2, 2, 14 are added to them respectively, the resulting numbers are in GP, we have to find the numbers.
We can now assume the three number as,
\[a-d,a,a+d\]
We are given that the sum of three number is 21, we can write it as,
\[\begin{align}
& \Rightarrow a=d+a+a-d=21 \\
& \Rightarrow a=7 \\
\end{align}\]
We can now substitute the value of a in the assumed three numbers, we get
\[\Rightarrow 7-d,7,7+d\] …….. (1)
We can now add the given three numbers, 2, 2, 14 in the above numbers, we get
\[\begin{align}
& \Rightarrow {{t}_{1}}=7+2-d=9-d \\
& \Rightarrow {{t}_{2}}=7+2=9 \\
& \Rightarrow {{t}_{3}}=7+14+d=21+d \\
\end{align}\]
We can see that the above terms are in GP.
We know that in GP, we have
\[\dfrac{{{t}_{2}}}{{{t}_{3}}}=\dfrac{{{t}_{1}}}{{{t}_{2}}}\]
We can now substitute the values in the above step, we get
\[\Rightarrow \dfrac{9}{21+d}=\dfrac{9-d}{9}\]
We can now simplify the above step by cross multiplication, we get
\[\begin{align}
& \Rightarrow 81=\left( 21+d \right)\left( 9-d \right) \\
& \Rightarrow {{d}^{2}}+12d-108=0 \\
& \Rightarrow \left( d+18 \right)\left( d-6 \right)=0 \\
\end{align}\]
We can now get the value of d,
\[d=-18,6\]
Therefore, the numbers in AP are
when d = -18, then from (1)
\[\Rightarrow 25,7,-11\]
When d = 6, then from (1), we get
\[\Rightarrow 1,7,13\]
Note: We should always remember that when the numbers are in GP, then we have some identities to solve these types of problems. We should first find the value of a from the given data to solve for d and find the numbers using the geometrical progression identities.
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