
If \[a,b,c\] are in A.P., then \[\frac{1}{{bc}},\frac{1}{{ca}},\frac{1}{{ab}}\] will be in
A. A.P.
B. G.P.
C. H.P.
D. None of these
Answer
161.7k+ views
Hint:
In this question, we need to find the type of another sequence given one sequence of arithmetic progression (A.P). For this, we must perform any one of the arithmetic operations, such as addition, subtraction, multiplication, or division, with the same number for all of A.P's numbers, and then simplify further to obtain the required solution.
Formula used:
The general arithmetic progression takes the form \[a,a + d,a + 2d,.....\]
The arithmetic progression's n-th terms is defined as
\[{a_n} = a + (n - 1)d\]
Complete step-by-step solution:
The general arithmetic progression takes the form \[a,a + d,a + 2d,.....\] where a is the first term and nth d is the common difference, which is the same distance between any two numbers in sequence; otherwise, the fixed number that must be added to any term of an AP to get the next term is known as the AP's common difference.
The arithmetic progression's n-th terms is defined as
\[{a_n} = a + (n - 1)d\]
Subtracting the first term from the second term, the second term from the third term, and so on yields the common difference ‘'d ‘'.
Now consider the following question:
If \[a,b,c\] are in A.P ---- (1)
We know that if we divide in A.P numbers by the same number, the resulting numbers will be A.P as well.
Now we have to divide the given sequence (1) by \[abc\]:
That implies, \[\frac{a}{{abc}},\frac{b}{{abc}},\frac{c}{{abc}}\] are in A.P
Now, by simplification,
We get, \[\frac{1}{{bc}},\frac{1}{{ac}},\frac{1}{{ab}}\]are in A.P
Therefore, if \[a,b,c\] are in A.P., then \[\frac{1}{{bc}},\frac{1}{{ca}},\frac{1}{{ab}}\] will be in A.P
Hence, the option A is correct.
Note:
Remember that the common difference (an addition or subtraction) between any two consecutive terms of an arithmetic sequence is a constant or the same. If we divide or multiply all the numbers in A.P by the same number, the resulting numbers in the sequence will also be in A.P.
In this question, we need to find the type of another sequence given one sequence of arithmetic progression (A.P). For this, we must perform any one of the arithmetic operations, such as addition, subtraction, multiplication, or division, with the same number for all of A.P's numbers, and then simplify further to obtain the required solution.
Formula used:
The general arithmetic progression takes the form \[a,a + d,a + 2d,.....\]
The arithmetic progression's n-th terms is defined as
\[{a_n} = a + (n - 1)d\]
Complete step-by-step solution:
The general arithmetic progression takes the form \[a,a + d,a + 2d,.....\] where a is the first term and nth d is the common difference, which is the same distance between any two numbers in sequence; otherwise, the fixed number that must be added to any term of an AP to get the next term is known as the AP's common difference.
The arithmetic progression's n-th terms is defined as
\[{a_n} = a + (n - 1)d\]
Subtracting the first term from the second term, the second term from the third term, and so on yields the common difference ‘'d ‘'.
Now consider the following question:
If \[a,b,c\] are in A.P ---- (1)
We know that if we divide in A.P numbers by the same number, the resulting numbers will be A.P as well.
Now we have to divide the given sequence (1) by \[abc\]:
That implies, \[\frac{a}{{abc}},\frac{b}{{abc}},\frac{c}{{abc}}\] are in A.P
Now, by simplification,
We get, \[\frac{1}{{bc}},\frac{1}{{ac}},\frac{1}{{ab}}\]are in A.P
Therefore, if \[a,b,c\] are in A.P., then \[\frac{1}{{bc}},\frac{1}{{ca}},\frac{1}{{ab}}\] will be in A.P
Hence, the option A is correct.
Note:
Remember that the common difference (an addition or subtraction) between any two consecutive terms of an arithmetic sequence is a constant or the same. If we divide or multiply all the numbers in A.P by the same number, the resulting numbers in the sequence will also be in A.P.
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