
If \[a,b,c\] are in A.P as well as in G.P. then
A. \[a = b \ne c\]
B. \[a \ne b = c\]
C. \[a \ne b \ne c\]
D. \[a = b = c\]
Answer
162.9k+ views
Hint: In the given question, we are given some arithmetic and geometric progression terms and must determine the relationship between the terms. For that, we must use the definitions of each term.
Formula used:
1.\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
2.\[ T_{_{n}}=a +(n-1)d\] where a= first term, d=common difference, = term
3.\[ a_{_{n}}=a r^{n-1}\] where a is the first term, r is the common ratio, and n is the number of the term.
4. If \[a,b,c\] are in A.P then \[2b=\,a+c\]
5. If \[a,b,c\] are in G.P then \[{b^2}=a\,c\]
Complete step-by-step solution:
According to the question, we are given that \[a,b,c\] are in A.P and we know that in arithmetic progression, the common difference between consecutive terms is the same. Therefore, we can write as
\[2b = \,a + c\,\,...\left( 1 \right)\]
Also, we know that the series of numbers where the ratio of any two consecutive terms is the same is called geometric progression. Therefore, we can write as
\[{b^2} = a\,c\,...\left( 2 \right)\]
Now we multiply equation (2) by \[4\], we get
\[4{b^2} = 4a\,c\]
Now we rewrite the above equation as:
\[{\left( {2b} \right)^2} = 4\,a\,c\,...\left( 3 \right)\]
Now we substitute the value \[2b\] from equation (1) in equation (3), we get
\[{\left( {a + c} \right)^2} = 4ac\]
Now we apply an algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], we get
\[{a^2} + {c^2} + 2\,a\,c\, = \,4ac\]
By simplifying, we get
\[
{a^2} + {c^2}\, = 4ac - 2ac \\
{a^2} + {c^2}\, = \,2ac \\
{a^2} + {c^2}\, - 2ac = 0 \\
{\left( {a - c} \right)^2} = 0
\]
Further solving,
\[a = c\,...\left( 4 \right)\]
Now substitute the value of \[a\] from equation (4) in equation (1), we get
\[a = b = c\]
Hence, option (D) is correct
Note: When responding to such a question, we must take care to define each type of progression series involved. Also, we must be aware of the key terminology involved and act immediately in accordance with the definition, always attempting to provide the simplest response and working step by step to reduce the risk of making a mistake.
Formula used:
1.\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
2.\[ T_{_{n}}=a +(n-1)d\] where a= first term, d=common difference, = term
3.\[ a_{_{n}}=a r^{n-1}\] where a is the first term, r is the common ratio, and n is the number of the term.
4. If \[a,b,c\] are in A.P then \[2b=\,a+c\]
5. If \[a,b,c\] are in G.P then \[{b^2}=a\,c\]
Complete step-by-step solution:
According to the question, we are given that \[a,b,c\] are in A.P and we know that in arithmetic progression, the common difference between consecutive terms is the same. Therefore, we can write as
\[2b = \,a + c\,\,...\left( 1 \right)\]
Also, we know that the series of numbers where the ratio of any two consecutive terms is the same is called geometric progression. Therefore, we can write as
\[{b^2} = a\,c\,...\left( 2 \right)\]
Now we multiply equation (2) by \[4\], we get
\[4{b^2} = 4a\,c\]
Now we rewrite the above equation as:
\[{\left( {2b} \right)^2} = 4\,a\,c\,...\left( 3 \right)\]
Now we substitute the value \[2b\] from equation (1) in equation (3), we get
\[{\left( {a + c} \right)^2} = 4ac\]
Now we apply an algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], we get
\[{a^2} + {c^2} + 2\,a\,c\, = \,4ac\]
By simplifying, we get
\[
{a^2} + {c^2}\, = 4ac - 2ac \\
{a^2} + {c^2}\, = \,2ac \\
{a^2} + {c^2}\, - 2ac = 0 \\
{\left( {a - c} \right)^2} = 0
\]
Further solving,
\[a = c\,...\left( 4 \right)\]
Now substitute the value of \[a\] from equation (4) in equation (1), we get
\[a = b = c\]
Hence, option (D) is correct
Note: When responding to such a question, we must take care to define each type of progression series involved. Also, we must be aware of the key terminology involved and act immediately in accordance with the definition, always attempting to provide the simplest response and working step by step to reduce the risk of making a mistake.
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