Arithmetic Progressions

Arithmetic Progressions

Sequence

A succession of numbers formed and arranged in a definite order according to a certain definite rule is called a sequence.

The number occurring at the nth place of a sequence is called its n^th term or the general term, to be denoted by t_n. A sequence is said to be finite or infinite according the number of distinct terms in it is finite or infinite.

Series

By adding the terms of a sequence, we obtain a series. A series is said to be finite or infinite depending on the number of terms added is finite or infinite.

Progressions

Sequences which follow certain patterns are called progressions. If the terms of a progressions are added upto certain terms it forms a series.

Thus, 1, 3, 5, 7, 9 ....; 8, 5, 2, -1, -4 ....

1, 2, 4, 8, 16 ....; 8, 4, 2, 1, 1/2 .... are each a sequence.

The n^th term of a sequence is usually denoted by T_n, where n = 1, 2, 3.

Thus T_l = first term, T₂ = second term, T₁₀ = tenth term and so on.

El. Write down the first five terms of the sequence with T_n = 4n – 1 / 5n+3

Sol. Put n = 1, 2, 3, 4, 5 in and find T₁, T₂, T₃, T₄, T₅ respectively.

Thus, the sequence is 3/8, 7/13, 11/18, 15/23, 19/28.

E2.   Show that the progression 7, 2, - 3, - 8, .... is an A.P. Find its 16th term and the general term.

Sol. Since (2 - 7) = (- 3 - 2) = [- 8- (- 3)] = - 5, it follows that the difference between any two consecutive terms is constant. So, it is an A.P.

Its first term = 7 and common difference = - 5.

16^th term=7+(16-1)x(-5)=-68.

And, n^th term = 7 + (n - 1) x (-5) = (12- 5n).

E3.   How many terms are there in the A.P. 7, 13, 19, 25, .... 205?

Sol. Let the given A.P. contains n terms. Then, clearly the nth term is 205.

.•. 7+(n-1) x 6=205

n=34.

Thus, the given A.P. contains 34 terms.

Some Important Results

1. If a fixed number is added (or subtracted) to each term of a given A.P., then the resulting sequence is also an A.P. and it has same common difference as that of the given A.P. e.g. 1, 4, 7, 10 .... is an A.P. If we add 4 to all the terms of the A.P., then the new sequence will be 5, 8, 11, 14, ..., so it is clear that the series is also an A.P. with the same common difference of 3.

2. If each term of an A.P. is multiplied by a constant (or divided by a non-zero constant), then the resulting sequence also an A.P. like 1, 8, 15, 22 is an A.P. If we multiply each term by 2 then the new series will be 2, 16, 30, 44 .... so the resulting sequence is also an A.P. with a common difference of 14.

3.   If a_l, a₂, a₃ .... and b_l, b₂, b₃, .... are two arithmetic progressions, then a_l + b_l, a₂ + b₂, a₃ + b₃, .... is also an AP. For example, if we take two A.Ps

(1)       1, 5, 9, 13 .... and

(2)       5, 7, 9, 11 ....

Now if we add the corresponding terms of the, A.P. then new sequence will be (1 + 5), (5 + 7), (9 + 9), (13 + 11) .... = 6, 12, 18, 24..... Thus, this sequence is also an A.P. with a common difference of 6.

Selection of terms in an A.P.

Sometimes we require certain number of terms in A.P. The following ways of selecting terms are generally very convenient.

Number		Common
	Terms
of terms		difference
3	a—d,a,a+d	d
4	a-3d,a—d,a+d,a+3d	2d

It should be noted that in case of an odd number of terms, the middle term is a and the common difference is d while in case of an even number of terms the middle terms are a — d, a + d and the common difference is 2d. The following examples will illustrate the use of such representations.

E4. The sum of three numbers in A.P. is —3 and their product is 8. Find the numbers.

Sol. Let the numbers be (a — d), a, (a + d). Then,

Sum =-3= (a—d)+a+(a+d)=-3

3a=-3

a = —1.

Product = 8

= (a—d)(a)(a+d)=8.

a(a² — d²) = 8 = (-1)(1 — d²) = 8 [ .• a = —1]

d²=9 d = ± 3.

If d = 3, the numbers are — 4, — 1, 2.

If d = — 3, the numbers are 2, —1, —4.

So, the numbers are — 4, — 1, 2 or 2, — 1, — 4.

E5. Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

        Sol. Let the numbers  be (a — 3d), (a — d), (a + d), (a + 3d). Then sum = 20

= (a-3d)+(a—d)+(a+d)+(a+3d)=20

4a=20.

a=5.

Sum of the squares = 120

= (a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 120

 a²-6da + 9d² + a²-2ad + d²+ a² + 2ad + d² + a² + 6ad + 9d² = 120

= 4a²+20d²= 120

a² + 5d² = 30

=25+5d²=30 [. a=5]

=5d²=5

 d=+1.

If d = 1, then the numbers are 2, 4, 6, 8.

If d = — 1, then the numbers are 8, 6, 4, 2.

Thus, the numbers are 2, 4, 6, 8 or 8, 6, 4, 2.

E5.  Find the second term of an A.P. whose sum of first three terms is 30.

Sol. Let the three terms be a — d, a, a + d. Then the sum of the first three terms = 30.

Therefore a - d + a + a + d = 3a

Given 3a = 30 = a = 10.

So the second term is 10.

E7. Find 8 arithmetic means between 1 and 10.

      Sol. Let the arithmetic means be A_I, A₂..... A₈ Then 1 A₁ , A₂ .... A₇, A₈, 10 will be an A. P. where 1 will be   first term and 10 will be 10^th term. So a+(10-1)d=10

    a+9d=10

   9d=9

    d=1.

   So the arithmetic means will be 2, 3, 4, 5, 6, 7, 8, 9.