Arithmetic progression(AP) or arithmetic sequence is a sequence of numbers in which each term after the first is obtained by adding a constant, d to the preceding term. The constant d is called common difference.

where a = the first term , d = the common difference

**Examples**

1, 3, 5, 7, ... is an arithmetic progression (AP) with a = 1 and d = 2

7, 13, 19, 25, ... is an arithmetic progression (AP) with a = 7 and d= 6

**n ^{th} term of an arithmetic progression**

_{n}= a + (n – 1)d

where t

_{n}= n

^{th}term, a= the first term , d= common difference

**Example 1**

Find 10^{th} term in the series 1, 3, 5, 7, ...

a = 1

d = 3 – 1 = 2

10^{th} term, t_{10} = a + (n-1)d = 1 + (10 – 1)2 = 1 + 18 = 19

**Example 2**

Find 16^{th} term in the series 7, 13, 19, 25, ...

a = 7

d = 13 – 7 = 6

16^{th} term, t_{16} = a + (n-1)d = 7 + (16 – 1)6 = 7 + 90 = 97

**Number of terms of an arithmetic progression**

where n = number of terms, a= the first term , l = last term, d= common difference

**Example**

Find the number of terms in the series 8, 12, 16, . . .72

a = 8

l = 72

d = 12 – 8 = 4

@@n=\dfrac{(l-a)}{d}+1=\dfrac{(72-8)}{4}+1\\=\dfrac{64}{4}+1=16+1=17@@

**Sum of first n terms in an arithmetic progression**

where a = the first term,

d= common difference,

@@l@@ = t

_{n}= n

^{th}term = a + (n-1)d

**Example 1**

Find 4 + 7 + 10 + 13 + 16 + . . . up to 20 terms

a = 4

d = 7 – 4 = 3

Sum of first 20 terms, S_{20}

@@ = \dfrac{n}{2} [2a+(n-1)d]\\= \dfrac{20}{2} [(2 \times 4)+(20-1)3]\\= 10 [8 + (19 \times 3)]\\=10(8+57)\\=650@@

**Example 2**

Find 6 + 9 + 12 + . . . + 30

a = 6

l = 30

d = 9 – 6 = 3

@@n = \dfrac{(l - a)}{d} + 1 \\= \dfrac{(30 - 6)}{3} + 1 \\= \dfrac{24}{3} + 1 \\= 8 + 1 \\= 9@@

Sum, S

@@=\dfrac{n}{2}(a+l)\\= \dfrac{9}{2}(6+30)\\=\dfrac{9}{2} \times 36 \\= 9 \times 18 \\= 162@@

**Arithmetic Mean**

If a, b, c are in AP, b is the Arithmetic Mean (AM) between a and c. In this case, @@ b = \dfrac{1}{2}(a + c)@@

_{1}, a

_{2}... a

_{n}, b are in AP we can say that a

_{1}, a

_{2}... a

_{n}are the n Arithmetic Means between a and b.

To solve most of the problems related to AP, the terms can be conveniently taken as

3 terms: (a – d), a, (a +d)

4 terms: (a – 3d), (a – d), (a + d), (a +3d)

5 terms: (a – 2d), (a – d), a, (a + d), (a +2d)

T_{n} = S_{n} - S_{n-1}

If each term of an AP is increased, decreased , multiplied or divided by the same non-zero constant, the resulting sequence also will be in AP.

In an AP, sum of terms equidistant from beginning and end will be constant.

Non-zero numbers @@a_1,\ a_2,\ a_3,\ \cdots \ a_n@@ are in Harmonic Progression(HP) if @@\dfrac{1}{a_1},\ \dfrac{1}{a_2},\ \dfrac{1}{a_3},\ \cdots \dfrac{1}{a_n}@@ are in AP. Harmonic Progression is also known as harmonic sequence.

**Examples**

@@\dfrac{1}{2}, \dfrac{1}{6}, \dfrac{1}{10}, \cdots @@ is a harmonic progression (HP)

^{th}term of the AP = a + (n - 1)d

Hence, if @@\dfrac{1}{a}, \dfrac{1}{a+d}, \dfrac{1}{a+2d}, \cdots@@ are in HP, n

^{th}term of the HP = @@\dfrac{1}{a + (n - 1)d}@@

In this case, @@ b = \dfrac{2ac}{a + c}@@

_{1}, a

_{2}... a

_{n}, b are in HP we can say that a

_{1}, a

_{2}... a

_{n}are the n Harmonic Means between a and b.

Geometric Progression(GP) or Geometric Sequence is sequence of non-zero numbers in which the ratio of any term and its preceding term is always constant.

^{2}, ar

^{3}, ...

where a = the first term , r = the common ratio

**Examples**

1, 3, 9, 27, ... is a geometric progression(GP) with a = 1 and r = 3

2, 4, 8, 16, ... is a geometric progression(GP) with a = 2 and r = 2

^{2}= ac

**n ^{th} term of a geometric progression(GP)**

where t

_{n}= n

^{th}term, a= the first term , r = common ratio, n = number of terms

**Example 1**

Find the 10^{th} term in the series 2, 4, 8, 16, ...

a = 2, r = @@\dfrac{4}{2}@@ = 2, n = 10

10^{th} term, t_{10}

@@=ar^{n-1} = 2 \times 2^{10-1} \\= 2 \times 2^{9} = 2 \times 512 = 1024@@

**Example 2**

Find 5^{th} term in the series 5, 15, 45, ...

a = 5, r = @@\dfrac{15}{5}@@ = 3, n = 5

5^{th} term, t_{5}

@@=ar^{n-1} = 5 \times 3^{5-1}\\= 5 × 3^4 = 5 × 81 = 405@@

**Sum of first n terms in a geometric progression(GP)**

where a= the first term,

r = common ratio,

n = number of terms

**Example 1**

Find 4 + 12 + 36 + ... up to 6 terms

a = 4, r = @@\dfrac{12}{4}@@ = 3, n = 6

Here r > 1. Hence,

@@S_6= \dfrac{a(r^n - 1)}{r - 1} = \dfrac{4(3^6 - 1)}{3 - 1} \\= \dfrac{4(729 - 1)}{2}= \dfrac{4 \times 728}{2} \\= 2 \times 728 = 1456@@

**Example 2**

Find @@1 + \dfrac{1}{2} + \dfrac{1}{4} +@@ ... up to 5 terms

a = 1, r = @@\dfrac{\left(\dfrac{1}{2}\right)}{1} = \dfrac{1}{2},@@ n = 5

Here r < 1. Hence,

@@S_6 = \dfrac{a(1 - r^n)}{1 - r} = \dfrac{1\left[1 - \left(\dfrac{1}{2}\right)^5 \right]}{\left(1 - \dfrac{1}{2}\right)} \\= \dfrac{\left(1 - \dfrac{1}{32} \right)}{\left(\dfrac{1}{2}\right)} = \dfrac{\left(\dfrac{31}{32}\right) }{\left(\dfrac{1}{2}\right)} = \dfrac{31}{16} = 1\dfrac{15}{16}@@

**Sum of an infinite geometric progression(GP)**

where a= the first term , r = common ratio

**Example**

Find @@1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}+\cdots \infty@@

a = 1, r = @@\dfrac{\left(\dfrac{1}{2}\right)}{1} = \dfrac{1}{2}@@

Here -1 < r < 1. Hence,

@@S_\infty = \dfrac{a}{1 - r} = \dfrac{1}{\left(1 - \dfrac{1}{2}\right)} = \dfrac{1}{\left(\dfrac{1}{2}\right)} = 2@@

**Geometric Mean**

If three non-zero numbers a, b, c are in GP, b is the Geometric Mean(GM) between a and c. In this case, @@ b = \sqrt{ac}@@

(Note that if a and b are of opposite sign, their GM is not defined.)

To solve most of the problems related to GP, the terms of the GP can be conveniently taken as

3 terms: @@\dfrac{a}{r}@@, a, ar

5 terms: @@\dfrac{a}{r^2}@@, @@\dfrac{a}{r}@@, a, ar, ar^{2}

If a, b, c are in GP, @@\dfrac{a-b}{b-c} = \dfrac{a}{b}@@

In a GP, product of terms equidistant from beginning and end will be constant.

GM

^{2}= AM × HM

Three non-zero numbers a, b and c are in HP if @@b = \dfrac{2ac}{a + c}@@

Three non-zero numbers a, b and c are in HP if @@\dfrac{a - b}{b - c} = \dfrac{a}{c}@@

(1) A > G > H

(2) A, G and H are in GP

@@1 + 1 + 1 + \cdots\text{ n terms}@@ @@= \sum{1} = n@@

@@1 + 2 + 3 + \cdots + n@@ @@=\sum{n}= \dfrac{n(n+1)}{2}%@@

@@1^2 + 2^2 + 3^2 + \cdots + n^2@@ @@= \sum{n^2}= \dfrac{n(n+1)(2n + 1)}{6}%@@

@@1^3 + 2^3 + 3^3 + \cdots + n^3@@ @@= \sum{n^3}= \dfrac{n^2(n+1)^2}{4} = \left[\dfrac{n(n+1)}{2}\right]^2 %@@

please tell me the formula

Under the section "5. Some Interesting Properties to Note", I observed you wrote HP rather than GP here

Three non-zero numbers a, b and c are in HP if (a−b)/(b-c) = a/c

Suppose (a−b)/(b-c) = a/c

=> (a-b)c = (b-c)a

=> ac - bc = ab - ac

=> 2ac = ab + bc

=> 2/b = 1/c + 1/a (divided all terms by abc)

i.e., 1/a , 1/b, 1/c are in AP

i.e., a,b,c are in HP

For instance consider the following incomplete sequence.

2,4,8,?

You can see that each term is twice it's predecessor. So, the unknown number is twice it's predecessor(2*8=16). Unknown number is 16.

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