
Identify the type of the following series.
\[1 + 2.2 + {3.2^2} + ... + {100.2^{99}}\]
A) AP
B) GP
C) HP
D) AGP
Answer
598.5k+ views
Hint: Arithmetic-Geometric Progression (AGP): This is a sequence in which each term consists of the product of an arithmetic progression and a geometric progression. In variables, it looks like,
\[a,(a + d)r,(a + 2d){r^2},...[a + (n - 1)d]{r^{n - 1}}\]
Complete step by step answer:
To find whether the type of the given series, we need to find out which form of sequence it will satisfy.
It is given that,
\[1 + 2.2 + {3.2^2} + ... + {100.2^{99}}\]
If we consider the above given series as a sequence, we get,
\[1,2.2,{3.2^2},...,{100.2^{99}}\]
From the above we conclude that,
1st term \[ = 1\]
2nd term \[ = 2.2\]
3rd term \[ = {3.2^2}\]
And the last term \[ = {100.2^{99}}\]
Now to find the type of series we should compare it with the general form of an arithmetic geometric progression.
By comparing with the general form of an arithmetic geometric progression we get,
\[a = 1\], \[d = 1\], \[r = 2\] and \[n = 100\]
As the given sequence is the product of Arithmetic progression (with\[a = 1\]&\[d = 1\]) and Geometric progression with (\[a = 1\] and \[r = 2\]) we can come to the following conclusion.
The given sequence satisfies the form of arithmetic geometric progression sequence.
Hence, we can come to a conclusion that the given series is an arithmetic geometric progression series.
The correct option is (D) AGP.
Note:
The general form of arithmetic progression is,
\[a,(a + d),(a + 2d),....[a + (n - 1)d]\], where, \[a\] be the initial term and \[d\] be the common difference.
Again, the general form of geometric progression is,
\[a,ar,a{r^2},...,a{r^{n - 1}}\], where, \[a\]be the initial term and \[r\] be the common ratio.
Since both general forms do not fit for the given series it is not considered. But initially we have to compare the given series with the general forms of AP and GP. Since HP is just the inverse of AP it is not taken into account.
\[a,(a + d)r,(a + 2d){r^2},...[a + (n - 1)d]{r^{n - 1}}\]
Complete step by step answer:
To find whether the type of the given series, we need to find out which form of sequence it will satisfy.
It is given that,
\[1 + 2.2 + {3.2^2} + ... + {100.2^{99}}\]
If we consider the above given series as a sequence, we get,
\[1,2.2,{3.2^2},...,{100.2^{99}}\]
From the above we conclude that,
1st term \[ = 1\]
2nd term \[ = 2.2\]
3rd term \[ = {3.2^2}\]
And the last term \[ = {100.2^{99}}\]
Now to find the type of series we should compare it with the general form of an arithmetic geometric progression.
By comparing with the general form of an arithmetic geometric progression we get,
\[a = 1\], \[d = 1\], \[r = 2\] and \[n = 100\]
As the given sequence is the product of Arithmetic progression (with\[a = 1\]&\[d = 1\]) and Geometric progression with (\[a = 1\] and \[r = 2\]) we can come to the following conclusion.
The given sequence satisfies the form of arithmetic geometric progression sequence.
Hence, we can come to a conclusion that the given series is an arithmetic geometric progression series.
The correct option is (D) AGP.
Note:
The general form of arithmetic progression is,
\[a,(a + d),(a + 2d),....[a + (n - 1)d]\], where, \[a\] be the initial term and \[d\] be the common difference.
Again, the general form of geometric progression is,
\[a,ar,a{r^2},...,a{r^{n - 1}}\], where, \[a\]be the initial term and \[r\] be the common ratio.
Since both general forms do not fit for the given series it is not considered. But initially we have to compare the given series with the general forms of AP and GP. Since HP is just the inverse of AP it is not taken into account.
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