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All the terms in A.P whose first term is 'a' and common difference d is squared. A different series is thus formed. This series is a?
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All the terms in A.P whose first term is 'a' and common difference d i...
When the term of an A. P are squared, no specific type of sequence is obtained.
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All the terms in A.P whose first term is 'a' and common difference d i...
Arithmetic Progression (AP)
An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. The first term of an AP is denoted by 'a', and the common difference is denoted by 'd'. The general form of an AP is:
a, a + d, a + 2d, a + 3d, ...

Squaring the Terms of an AP
When we square each term of an AP, we obtain a new series. Let's denote this new series as 'S'. Each term in 'S' will be the square of the corresponding term in the original AP. Therefore, the first term of 'S' will be (a^2), the second term will be ((a + d)^2), the third term will be ((a + 2d)^2), and so on.

Deriving the General Form of the Squared Series
To find a general form for the squared series, we can expand the squares of the terms in the original AP.

The squared series will be:
(a^2), (a^2 + 2ad + d^2), (a^2 + 4ad + 4d^2), (a^2 + 6ad + 9d^2), ...

Identifying the Pattern
Upon observing the squared series, we can notice a pattern:

The first term, (a^2), is the square of the first term in the original AP.
The second term, (a^2 + 2ad + d^2), can be written as (a + d)^2.
The third term, (a^2 + 4ad + 4d^2), can be written as (a + 2d)^2.
The fourth term, (a^2 + 6ad + 9d^2), can be written as (a + 3d)^2.

From this pattern, we can conclude that the squared series is also an AP, with a first term of (a^2) and a common difference of (2ad + d^2).

Conclusion
In summary, when we square each term of an AP, we obtain a new series that is also an AP. The first term of this squared series is (a^2), and the common difference is (2ad + d^2).
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All the terms in A.P whose first term is 'a' and common difference d is squared. A different series is thus formed. This series is a?
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