Arithmetic Progression-白红宇

Arithmetic Progression

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发布时间：2019-05-24

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Arithmetic Progression (A.P.)

Let us recall some formulae and properties studied earlier.

A sequence a1, a2, a3,…, an,… is called arithmetic sequence or arithmetic progression if an + 1 = an+ d, n ∈ N, where a1 is called the first term and the constant term d is called the common difference of the A.P.

Let us consider an A.P. (in its standard form) with first term a and common difference d, i.e., a, a + d, a + 2d, ...

Then the nth term (general term) of the A.P. is an = a + (n – 1) d.

We can verify the following simple properties of an A.P. :

(i) If a constant is added to each term of an A.P., the resulting sequence is also an A.P.

(ii) If a constant is subtracted from each term of an A.P., the resulting sequence is also an A.P.

(iii) If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

(iv) If each term of an A.P. is divided by a non-zero constant then the resulting sequence is also an A.P.

Here, we shall use the following notations for an arithmetic progression:

a = the first term, l = the last term, d = common difference,

n = the number of terms.

Sn= the sum to n terms of A.P.

Let a, a + d, a + 2d, …, a + (n – 1) d be an A.P. Then

l = a + (n – 1) d

Sn = n/2[2a + (n-1)d]

We can also write, Sn = n/2[a + l]

Let us consider some examples.

Example In an A.P. if mth term is n and the nth term is m, where m ≠ n, find the pth term.

Solution We have am = a + (m – 1) d = n, ... (1)

And an = a + (n – 1) d = m ... (2)

Solving (1) and (2), we get

(m – n) d = n – m, or d = – 1, ... (3)

and a = n + m – 1 ... (4)

Therefore ap= a + (p – 1)d

= n + m – 1 + ( p – 1) (–1) = n + m – p

Hence, the pth term is n + m – p.

Example If the sum of n terms of an A.P. is nP + ½ n(n – 1)Q , where P and Q are constants, find the common difference.

Solution Let a1, a2, … an be the given A.P. Then

Sn = a1 + a2 + a3 +...+ an-1 + an = nP + ½ n (n – 1) Q

Therefore S1 = a1 = P, S2 = a1 + a2 = 2P + Q

So that a2 = S2 – S1 = P + Q

Hence, the common difference is given by d = a2 – a1 = (P + Q) – P = Q.

Example The income of a person is Rs. 3,00,000, in the first year and he receives an increase of Rs.10,000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.

Solution Here, we have an A.P. with a = 3,00,000, d = 10,000, and n = 20.

Using the sum formula, we get,

S20 = 20/2 [600000 + 19 x 10000] = 10 (790000) = 79,00,000.

Hence, the person received Rs. 79,00,000 as the total amount at the end of 20 years.

Arithmetic mean

Given two numbers a and b. We can insert a number A between them so that a, A, b is an A.P. Such a number A is called the arithmetic mean (A.M.) of the numbers a and b. Note that, in this case, we have

A – a = b – A, i.e., A = (a + b) / 2

We may also interpret the A.M. between two numbers a and b as their average (a + b) / 2. For example, the A.M. of two numbers 4 and 16 is 10. We have, thus constructed an A.P. 4, 10, 16 by inserting a number 10 between 4 and 16. The natural question now arises : Can we insert two or more numbers between given two numbers so that the resulting sequence comes out to be an A.P. ? Observe that two numbers 8 and 12 can be inserted between 4 and 16 so that the resulting sequence 4, 8, 12, 16 becomes an A.P. More generally, given any two numbers a and b, we can insert as many numbers as we like between them such that the resulting sequence is an A.P.

Let A1, A2, A3, …, An be n numbers between a and b such that a, A1, A2, A3, …, An, b is an A.P.

Here, b is the (n + 2)th term, i.e., b = a + [(n + 2) – 1]d = a + (n + 1) d. This gives d = (b – a)/(n + 1).

Thus, n numbers between a and b are as follows: