Mathematics SEQUENCES AND SERIES - Arithmetic Progression (A.P.) and Arithmetic mean
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star Arithmetic Progression (A.P.)
star Arithmetic mean

### Arithmetic Progression (A.P.)

\color{fuchsia} \(★ mathbf\ul"Arithmetic Progression (A.P.)")

A sequence color(blue)(a_1, a_2, a_3,…..., a_n,…...) is called arithmetic sequence or arithmetic progression

if color(red)(a_(n + 1) = a_n + d) , n ∈ N, where color(blue)(a_1) is called color(blue)("the first term") and the constant term color(blue)(d) is called color(blue)("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 n^(th) term (color(blue)("general term")) of the A.P.  is color(red)(a_n = a + (n – 1) d.)

We can verify the following color(green)("simple properties of an A.P. :")
color(green)((i)) If a constant is added to each term of an A.P., the resulting sequence is also an A.P.

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

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

color(green)((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 color(red)("following notations for an arithmetic progression:")

color(blue)(a) = the first term,  \ \ \ \ \ \color(blue)( l )= the last term,  \ \ \ \ \ color(blue)(d) = common difference,

color(blue)(n) = the number of terms.

color(blue)(S_n)= the sum to n terms of A.P.

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

color(red)(S_n=(n/2)[2a+(n-1)d])

We can also write, color(red)(S_n=(n/2)[a+l])
Q 3089412317

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 a_m = a + (m – 1) d = n, ... (1)
and a_n = 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 a_p= a + (p – 1)d
= n + m – 1 + ( p – 1) (–1) = n + m – p
Hence, the pth term is n + m – p.
Q 3019512410

If the sum of n terms of an A.P. is nP +1/2 n (n-1) Q where P and Q are constants, find the common difference.

Solution:

Let a_1, a_2, … a_n be the given A.P. Then
S_n = a_1 + a_2 + a_3 +...+ a_(n–1) + a_n = nP +1/2 n (n – 1) Q
Therefore S_1 = a_1 = P, S_2 = a_1 + a_2 = 2P + Q
So that a_2 = S_2 – S_1 = P + Q
Hence, the common difference is given by d = a_2 – a_1 = (P + Q) – P = Q.
Q 3019612510

The sum of n terms of two arithmetic progressions are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 12th terms.

Solution:

Let a_1, a_2 and d_1, d_2 be the first terms and common difference of the first and second arithmetic progression, respectively. According to the given condition, we have

text(Sum to terms of first A.P.)/text(Sum to terms of second A.P.)

 = (3n+8)/(7n+15)

or (n/2 [ 2a_1 +(n-1)d_1])/(n/2 [ 2a_2 + (n-1)d_2]) = (3n+8)/(7n+15)

or (2a_1+(n-1)d_1)/(2a_2+(n-1)d_2) = (3n+8)/(7n+15) .....(1)

Now text(12 term of first A.P.)/text(12 term of second A.P) = (a_1+11d_1)/(a_2+11d_2)

(2a_1+22d_1)/(2a_2+22d_2) = (3xx23+8)/(7xx23+15) [By putting n = 23 in (1)]

Therefore (a_1+11d_1)/(a_2+11d_2) = text(12 term of first A.P.)/text(12 term of second A.P.)= 7/16

Hence, the required ratio is 7 : 16.
Q 3029612511

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, S_(20) = 20/2 [600000+ 19 xx10000]

 = 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

\color{fuchsia} \(★ mathbf\ul"Arithmetic mean")

\color{red} ✍️ Given two numbers a and b. We can insert a number A between them so that a, A, b is an A.P.

\color{red} ✍️ Such a number color(blue)(A) is called the color(blue)"arithmetic mean" (A.M.) of the numbers a and b.

\color{red} ✍️ Note that, in this case, we have A – a = b – A, i.e., A = (a + b)/2

\color{red} ✍️ We may also interpret the A.M. between two numbers a and b as their average ( a + b)/2

\color{blue} => 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 color(blue)(A_1, A_2, A_3, …, A_n) be n numbers between a and b such that color(blue)(a, A_1, A_2, A_3, …, A_n, 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:

A_1 = a + d = a +(b-a)/(n+1)

A_2 = a + 2d = a +2 * (b-a)/(n+1)

A_3 = a + 3d = a +3 * (b-a)/(n+1)

..... ..... ..... .....
..... ..... ..... .....

A_n = a + nd = a +n * (b-a)/(n+1)

.
Q 3009712618

Insert 6 numbers between 3 and 24 such that the resulting sequence is an A.P.

Solution:

Let A_1, A_2, A_3, A_4, A_5 and A_6 be six numbers between 3 and 24 such that 3, A_1, A_2, A_3, A_4, A_5, A_6, 24 are in A.P. Here, a = 3, b = 24, n = 8.
Therefore, 24 = 3 + (8 –1) d, so that d = 3.
Thus A_1 = a + d = 3 + 3 = 6; A_2 = a + 2d = 3 + 2 × 3 = 9;
A_3 = a + 3d = 3 + 3 × 3 = 12; A_4 = a + 4d = 3 + 4 × 3 = 15;
A_5 = a + 5d = 3 + 5 × 3 = 18; A_6 = a + 6d = 3 + 6 × 3 = 21.
Hence, six numbers between 3 and 24 are 6, 9, 12, 15, 18 and 21.