Formula:

The sum of an arithmetico geometric sequence is generally calculated by applying the method discussed below. It's formula is as follows,

However, this formula is seldom used, since it is too complex to remember (if you do remember it, great!). On the other hand, its derivation is a sequential process, and thus is applied whenever you have to find the sum of an arithmetico geometric sequence. An example of application of this derivation is given below.

Derivation:

Consider an arithmetico geometric sequence of 'n' terms as follows:

Moving (1 - r) from the LHS to the RHS,

This is a formula for the sum of 'n' terms of an arithmetico geometric progression, but since it is too complex and long to remember, we apply the whole method explained above in problems on the sum of arithmetico geometric sequences. In short, this method involves three steps:

An example:

The given arithmetico geometric sequence is formed from the arithmetic sequence

Therefore for the above arithmetico geometric sequence,

Substituting the values of a, d and r, we get:

Simplifying, we get

Now we can write the sum of the artihmetico geometric sequence as

Dividing both sides by

This is the sum of the given arithmetico geometric sequence.

The sum of an arithmetico geometric sequence is generally calculated by applying the method discussed below. It's formula is as follows,

However, this formula is seldom used, since it is too complex to remember (if you do remember it, great!). On the other hand, its derivation is a sequential process, and thus is applied whenever you have to find the sum of an arithmetico geometric sequence. An example of application of this derivation is given below.

Derivation:

Consider an arithmetico geometric sequence of 'n' terms as follows:

The sum of the above arithmetico geometric sequence isa, (a + d)r, (a + 2d)r^{2}, .... , (a + (n - 1)d)r^{(n - 1)}

Now multiply the above sum with constant ratio 'r',Sn = a + (a + d)r + (a + 2d)r^{2}+ .... + (a + (n - 1)d)r^{(n - 1)}

Now subtract the above two equations, so we obtainr Sn = ar + (a + d)r^{2}+ (a + 2d)r^{3}+ .... + (a + (n - 2)d)r^{(n - 1)}+ (a + (n - 1)d)r^{n}

Notice that in the above equation, the terms from dr till dr^(n - 1) form a geometric progression with first term 'dr' and common ratio 'r'. Therefore we can rewrite the above equation asSn(1 - r) = a + dr + dr^{2}+ .... + dr^{(n - 1)}- (a + (n - 1)d)r^{n}

Moving (1 - r) from the LHS to the RHS,

This is a formula for the sum of 'n' terms of an arithmetico geometric progression, but since it is too complex and long to remember, we apply the whole method explained above in problems on the sum of arithmetico geometric sequences. In short, this method involves three steps:

- Write the sum of the arithmetico geometric sequence as a series
- Multiply the above equation with 'r', the common ratio
- Subtract the above two equations

Now let us discuss an example of how the above method is applied in actual math problems to find the sum of arithmetico geometric sequences:

An example:

**: Find the sum of the following arithmetico geometric sequence:**

__Question__2, 5x, 8x, ... upto '^{2}, 11x^{3}n' terms

**:**__Solution__The given arithmetico geometric sequence is formed from the arithmetic sequence

*2, 5, 8, 11,*.. and the geometric sequence*1, x, x*..^{2}, x^{3}, .Therefore for the above arithmetico geometric sequence,

- First term of arithmetic sequence,
*a = 2* - Common difference of arithmetic sequence,
*d = 3* - Common ratio of geometric sequence,
*r = x*

*n*term of the above arithmetico geometric sequence is:

^{th}

**T**_{n}= (a + (n - 1)d)r^{(n - 1)}**T**_{n}= (2 + (n - 1)3)x^{(n - 1)}Simplifying, we get

*T*_{n}= (3n - 1)x^{(n - 1)}
and

That is,*(n - 1)*term of the above sequence is^{th}*T*_{n - 1}= (a + (n - 2)d)r^{n - 2}*T*_{n - 1}= (2 + (n - 2)3)x^{n - 2}= (3n - 4)x^{n - 2}Now we can write the sum of the artihmetico geometric sequence as

Multiplying the above sum by common ratio 'r',S_{n}= 2 + 5x + 8x^{2}+ 11x^{3}... + (3n - 1)x^{(n - 1)}

Subtracting the above two equations,x S_{n}= 2x + 5x^{2}+ 8x^{3}+ ... + (3n - 4)x^{n - 2}+ (3n - 1)x^{n}

In the above equation, the terms ranging fromS_{n}(1 - x) = 2 + 3x + 3x^{2}+ 3x^{3}+ ... + (3n - 4)x^{n - 1}- (3n - 1)x^{n}

*3x*to*(3n - 4)x*form a geometric series, therefore we can write the above equation as,^{n - 1}Dividing both sides by

*we obtain***(1 - x)**This is the sum of the given arithmetico geometric sequence.

Very nice, understandable derivation.

ReplyDeleteThnx Man (y)

ReplyDeleteThnx Man

ReplyDeleteGood Work !! Thanks.

ReplyDeleteGood Work !! Thanks.

ReplyDeleteThanks a LOTT

ReplyDeleteThanx alot

ReplyDeleteThank you.. I finally got it!! :)

ReplyDeletethanks alot

ReplyDeletewould have been better if you would have solved an example

anyways thanks :)

Thanks a lot... It is a clearly understandable derivation.

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ReplyDeleteThanks a lot!!

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