
Sequence 3,12,48,192,... is geometric or arithmetic, how to determine that?
Answer
255.6k+ views
Hint:The sequence whose terms are obtained by the multiplication of a number to its previous term called common ratio or multiplier. Such a sequence is the geometric sequence. The generalized form of terms of a geometric sequence is \[{{a}_{n}}={{a}_{0}}{{q}^{n-1}}\], this gives the nth term where \[{{a}_{0}}\] is the first term and q is the multiplier.
Complete step-by-step solution:
In the given question, the sequence is 3, 12, 48, 192, …
Now, let the first term,\[{{a}_{0}}\] = 3. The second term is \[{{a}_{1}}\]=12, third term is \[{{a}_{2}}\] = 48 and the fourth term is \[{{a}_{3}}\] = 192.
Now let us divide the second term with first term then we get
\[\dfrac{{{a}_{1}}}{{{a}_{0}}}\] = \[\dfrac{12}{3}\] which is equal to 4 ---(1)
Now we divide the third term by second term then we get
\[\dfrac{{{a}_{2}}}{{{a}_{1}}}\] = \[\dfrac{48}{12}\] which is also equal to 4 --(2)
Now we divide the fourth term by the third term then we get
\[\dfrac{{{a}_{3}}}{{{a}_{2}}}\] = \[\dfrac{192}{48}\] which is also equal to 4 --(3)
From equations (1), (2) and (3), we can say that the terms have a common ratio equal to 4 and hence it is a geometric sequence. And the nth term of this sequence is given by \[{{a}_{n}}=3{{(4)}^{n-1}}\].
Note: While solving questions from a geometric sequence, one common error would be not correctly finding the value of r, the common multiplier. Sometimes sequences of fractions are confusing. You might check that the r calculated is consistently true for any two successive terms of the sequence. This helps to verify the sequence.
Complete step-by-step solution:
In the given question, the sequence is 3, 12, 48, 192, …
Now, let the first term,\[{{a}_{0}}\] = 3. The second term is \[{{a}_{1}}\]=12, third term is \[{{a}_{2}}\] = 48 and the fourth term is \[{{a}_{3}}\] = 192.
Now let us divide the second term with first term then we get
\[\dfrac{{{a}_{1}}}{{{a}_{0}}}\] = \[\dfrac{12}{3}\] which is equal to 4 ---(1)
Now we divide the third term by second term then we get
\[\dfrac{{{a}_{2}}}{{{a}_{1}}}\] = \[\dfrac{48}{12}\] which is also equal to 4 --(2)
Now we divide the fourth term by the third term then we get
\[\dfrac{{{a}_{3}}}{{{a}_{2}}}\] = \[\dfrac{192}{48}\] which is also equal to 4 --(3)
From equations (1), (2) and (3), we can say that the terms have a common ratio equal to 4 and hence it is a geometric sequence. And the nth term of this sequence is given by \[{{a}_{n}}=3{{(4)}^{n-1}}\].
Note: While solving questions from a geometric sequence, one common error would be not correctly finding the value of r, the common multiplier. Sometimes sequences of fractions are confusing. You might check that the r calculated is consistently true for any two successive terms of the sequence. This helps to verify the sequence.
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