
How do you tell whether \[1,2,4,7,11\] is in arithmetic progression?
Answer
491.4k+ views
Hint: In this question, we have to find whether the given sequence is an arithmetic progression. This can be done by finding the common difference, we will use the common difference formula which is given by, Common difference \[(d)\] \[ = {a_{n + 1}} - {a_n}\]. So, if all the common differences are equal then we can say that they are in arithmetic progression.
Complete step by step solution:
An arithmetic progression is a sequence where the difference between every two consecutive terms is the same. An arithmetic progression is a sequence where each term, except the first term, is obtained by adding a fixed number to its previous term.
The general arithmetic progression is of the form \[a,a + d,a + 2d,...\] where \[a\] is the first term and \[d\] is the common difference. The \[{n^{th}}\] term of the arithmetic progression is defined as \[{a_n} = {a_0} + \left( {n - 1} \right)d\].
Now the given sequence is \[1,2,4,7,11\]. Here, \[{a_1} = 1,{\text{ }}{a_2} = 2,{\text{ }}{a_3} = 4,{\text{ }}{a_4} = 7{\text{ and }}{a_5} = 11\].
The difference between \[{a_1}\] and \[{a_2}\] is given by \[{a_2} - {a_1} = 2 - 1\] i.e., \[{a_2} - {a_1} = 1\].
The difference between \[{a_2}\] and \[{a_3}\] is given by \[{a_3} - {a_2} = 4 - 2\] i.e., \[{a_3} - {a_2} = 2\].
The difference between \[{a_3}\] and \[{a_4}\] is given by \[{a_4} - {a_3} = 7 - 4\] i.e., \[{a_4} - {a_3} = 3\].
The difference between \[{a_4}\] and \[{a_5}\] is given by \[{a_5} - {a_4} = 11 - 7\] i.e., \[{a_5} - {a_4} = 4\].
Here, we can see that the difference between a term and its immediately preceding term is not the same.
Therefore, \[1,2,4,7,11\] is not arithmetic.
Note:
\[1,2,4,7,11\] is also not a geometric sequence. Geometric sequence is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. But here \[\dfrac{2}{1} = 2\], \[\dfrac{4}{2} = 2\] but \[\dfrac{7}{4} \ne 2\]. Common ratio of all the terms is not equal. Therefore, it is also not a geometric sequence.
Complete step by step solution:
An arithmetic progression is a sequence where the difference between every two consecutive terms is the same. An arithmetic progression is a sequence where each term, except the first term, is obtained by adding a fixed number to its previous term.
The general arithmetic progression is of the form \[a,a + d,a + 2d,...\] where \[a\] is the first term and \[d\] is the common difference. The \[{n^{th}}\] term of the arithmetic progression is defined as \[{a_n} = {a_0} + \left( {n - 1} \right)d\].
Now the given sequence is \[1,2,4,7,11\]. Here, \[{a_1} = 1,{\text{ }}{a_2} = 2,{\text{ }}{a_3} = 4,{\text{ }}{a_4} = 7{\text{ and }}{a_5} = 11\].
The difference between \[{a_1}\] and \[{a_2}\] is given by \[{a_2} - {a_1} = 2 - 1\] i.e., \[{a_2} - {a_1} = 1\].
The difference between \[{a_2}\] and \[{a_3}\] is given by \[{a_3} - {a_2} = 4 - 2\] i.e., \[{a_3} - {a_2} = 2\].
The difference between \[{a_3}\] and \[{a_4}\] is given by \[{a_4} - {a_3} = 7 - 4\] i.e., \[{a_4} - {a_3} = 3\].
The difference between \[{a_4}\] and \[{a_5}\] is given by \[{a_5} - {a_4} = 11 - 7\] i.e., \[{a_5} - {a_4} = 4\].
Here, we can see that the difference between a term and its immediately preceding term is not the same.
Therefore, \[1,2,4,7,11\] is not arithmetic.
Note:
\[1,2,4,7,11\] is also not a geometric sequence. Geometric sequence is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. But here \[\dfrac{2}{1} = 2\], \[\dfrac{4}{2} = 2\] but \[\dfrac{7}{4} \ne 2\]. Common ratio of all the terms is not equal. Therefore, it is also not a geometric sequence.
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