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TheArithmetic Progression (A.P.) is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference (d).
For Example:
The Sequence:
2, 5, 8, 11, 14 ....
It is an AP with the common difference (d) = 3.
Arithmetic Progression (A.P.) properties simplify calculations and provide a structured way to analyze sequences. From understanding how terms relate to common differences, to deriving sums and relationships, the properties of A.P. form the foundation for solving many mathematical problems.
Properties of Arithmetic Progression (AP) with Examples:
Adding or subtracting the same constant from all terms results in another A.P.
Example:
3, 6, 9, 12, 15, . . . (A.P)
Add 2 to each term gives: (+2)
5, 8, 11, 14, 17, . . . (A.P)
Similarly, this property can be applied for subtraction also it will give the AP.
Proof:
Let x1, x2, x3, ⦠be an Arithmetic Progression (A.P.) with a common difference d. Let a fixed constant k be added to or subtracted from each term of the A.P. The new sequence becomes:
x1 + k, x2 + k, x3 + k . . . .
or
x1 - k, x2 - k, x3 - k . . . .
Let yn = xn + k, where n =1, 2, 3, . . .. The new sequence is y1, y2, y3, . . . ..
The difference between consecutive terms in the new sequence is:
yn + 1 - yn = (xn+1 + k) ā (xn + k)
yn + 1 - yn = (xn+1 - xn) + (k- k)
yn + 1 - yn = xn+1 - xnSince xn+1 - xn = d for all n ā N:
yn+1 - yn = d
The new sequence y1, y2, y3, is also an Arithmetic Progression, and the common difference d remains unchanged when a constant k is added to or subtracted from each term.
Multiplying or dividing all terms of an A.P. by the same constant results in another Arithmetic Progression (A.P.), where a common difference is scaled.
Example:
3, 6, 9, 12, 15, . . . (A.P.)
Multiply each term by k = 2: (Ć2)
6, 12, 18, 24, 30, . . . (A.P.)
Here, the common difference will be multiplied by the same constant k Ć d
Proof:
Let x1, x2, x3, . . . be an Arithmetic Progression (A.P.) with a common difference d. If each term of the A.P. is multiplied by a non-zero constant k, the new sequence becomes:
b1 = x1 ā k,āb2 = x2 ā k,āb3 = x3 ā k, . . .
To Prove: The resulting sequence b1, b2, b3, . . . is also an A.P., and its common difference is dā²= k ā d
The common difference of the new sequence is :
bn+1ā ā bn ā= (xn+1 āā k) ā (xn āā k)
bn+1 āā bnā = k ā (xn+1 āā xnā)Since xn+1 ā xn = d (the common difference of the original A.P.):
bn+1 ā bn = k ā d
Thus, the new sequence has a constant difference dā² = k ā d
If each term of an A.P. is multiplied by a constant k, the resulting sequence is also an A.P., with the common difference scaled to dā² = k ā d.
Note: If all terms of an AP are multiplied by a constant k, the resulting sequence is also an AP, with the common difference scaled to k ā d. Similarly, if all terms are divided by k, the resulting sequence is an AP with a common difference of d/k.
In a finite Arithmetic Progression (A.P.), the sum of terms equidistant from the beginning and end is always equal to the sum of the first and last terms.
In simpler terms, if you pick two terms with the same distance from the beginning and the end of the sequence, their sum will always be the same. This sum is equal to the sum of the first term and the last term.
Example:
Consider the finite A.P.:
2, 5, 8, 11, 14, 17,
Here, a1 = 2, an = 17, and n = 6
- Sum of the 1st and 6th terms:
- a1 + a6 = 2 + 17 = 19
- Sum of the 2nd and 5th terms:
- a2 + a5 = 5 + 14 = 19
- Sum of the 3rd and 4th terms:
- a3 + a4 = 8 + 11 = 19
If the Number of Terms is Odd:
If the number of Terms is Even:
Any three consecutive terms in an A.P. follow the property:
This means the middle term is the arithmetic mean of the terms surrounding it.
2y = x + z