`star` Arithmetic Progression (A.P.)

`star` Arithmetic mean

`star` Arithmetic mean

`\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])`

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.

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.

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.

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.

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.

`\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)`

.

`\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.

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.