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

180/3 + 4 = 64

64/4 + 5 = 21

21/5 + 6 = 10.2

10.2/6 + 7 =

**8.7**

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