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If a, b, c are in A.P and p is the A.M between a and b, and q is the A.M between b and c, show that b is the A.M between p and q.
Answer
496.5k+ views
Hint: In this question, as a, b, c are in A.P we get a relation between them as \[2b=a+c\]. Then using the A.M formula \[A=\dfrac{a+b}{2}\]we can get the values of p in terms of a and b, q in terms of b and c. Now, from the relation of a, b, c on further simplification we can get the relation between p and q in terms of b.
Complete step-by-step answer:
Now, from the given condition in the question we have a, b, c are in A.P
As we already know that if the terms a, b, c are in A.P then they satisfy the condition that
\[\Rightarrow 2b=a+c\]
ARITHMETIC MEAN (A.M):
If a, A, b are in A.P, then \[A=\dfrac{a+b}{2}\]is called the arithmetic mean of a and b.
Now, given that p is the A.M between a and b
Now, from the arithmetic mean formula we get,
\[\Rightarrow p=\dfrac{a+b}{2}\]
Now, also given that q is the arithmetic mean between b ad c
Now, from the above arithmetic mean formula we get
\[\Rightarrow q=\dfrac{b+c}{2}\]
Now, from the relation between a, b, c we have
\[\Rightarrow 2b=a+c\]
Now, this can also be written as
\[\Rightarrow c=2b-a\]
Let us now substitute this value of c in the value of q obtained
\[\Rightarrow q=\dfrac{b+2b-a}{2}\]
Now, from the value of p we have
\[\Rightarrow p=\dfrac{a+b}{2}\]
Now, this can also be written as
\[\Rightarrow a=2p-b\]
Let us now again substitute this value of a in the value of q above
\[\Rightarrow q=\dfrac{b+2b-\left( 2p-b \right)}{2}\]
Let us now multiply with 2 on both the sides
\[\Rightarrow 2q=b+2b-\left( 2p-b \right)\]
Now, on further simplification we get,
\[\Rightarrow 2q=4b-2p\]
Now, on rearranging the terms we get,
\[\Rightarrow 4b=2q+2p\]
Let us now divide with 4 on both sides
\[\therefore b=\dfrac{p+q}{2}\]
Hence, b is the arithmetic mean of p and q
Note:Instead of considering the relation between a, b, c we can also say that b is the arithmetic mean of a and c. Then on further simplification we can also say that p, b, q are in A.P which on solving further gives the answer. Both the methods give the same result.
It is important to note that we need to write the value of a and c in terms of b, p and q as we need to get the relation between those three.
Complete step-by-step answer:
Now, from the given condition in the question we have a, b, c are in A.P
As we already know that if the terms a, b, c are in A.P then they satisfy the condition that
\[\Rightarrow 2b=a+c\]
ARITHMETIC MEAN (A.M):
If a, A, b are in A.P, then \[A=\dfrac{a+b}{2}\]is called the arithmetic mean of a and b.
Now, given that p is the A.M between a and b
Now, from the arithmetic mean formula we get,
\[\Rightarrow p=\dfrac{a+b}{2}\]
Now, also given that q is the arithmetic mean between b ad c
Now, from the above arithmetic mean formula we get
\[\Rightarrow q=\dfrac{b+c}{2}\]
Now, from the relation between a, b, c we have
\[\Rightarrow 2b=a+c\]
Now, this can also be written as
\[\Rightarrow c=2b-a\]
Let us now substitute this value of c in the value of q obtained
\[\Rightarrow q=\dfrac{b+2b-a}{2}\]
Now, from the value of p we have
\[\Rightarrow p=\dfrac{a+b}{2}\]
Now, this can also be written as
\[\Rightarrow a=2p-b\]
Let us now again substitute this value of a in the value of q above
\[\Rightarrow q=\dfrac{b+2b-\left( 2p-b \right)}{2}\]
Let us now multiply with 2 on both the sides
\[\Rightarrow 2q=b+2b-\left( 2p-b \right)\]
Now, on further simplification we get,
\[\Rightarrow 2q=4b-2p\]
Now, on rearranging the terms we get,
\[\Rightarrow 4b=2q+2p\]
Let us now divide with 4 on both sides
\[\therefore b=\dfrac{p+q}{2}\]
Hence, b is the arithmetic mean of p and q
Note:Instead of considering the relation between a, b, c we can also say that b is the arithmetic mean of a and c. Then on further simplification we can also say that p, b, q are in A.P which on solving further gives the answer. Both the methods give the same result.
It is important to note that we need to write the value of a and c in terms of b, p and q as we need to get the relation between those three.
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