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there are many in progression who to know these pattern.can u give me all the pattern that i will come across.

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8 years ago SHAIK AASIF AHAMED
74 Points
```							Hello student,There are basically 4 types of progressionsThey are arithmetic progression,
If the first term of the Arithmetic Progression which is generally called as A.P. is ‘a’ and the common difference is ‘d’ then the nth term is given by tn = a+(n-1)d.A sequence of numbers <an> is said to be in arithmetic progression if the difference between the terms is constant i.e. tn – tn-1 is a constant for all n ∈ N. the constant difference between the terms is called the common difference and is denoted by‘d’.

Similarly, the sum of first ‘n’ terms of the A.P. is given by

Sn= n/2 [2a + (n-1) d]

= n/2 [a + l], where l is the last term of the A.P.
Geometric progression,
If the first term of the geometricprogression is ‘a’ and the common difference is ‘r’ then the nth term is given by tn = arn-1If every succesiveterm is a fixed multiple of the preceding term then the series is said to be a geometric series. A progression of the form a, ar, ar2, ….. is a geometric progression where the common multiple r is called the fixed ratio. it follows from the definition itself that no term of the G.P can be zero because in such a case the series will get reducedto a zero series.

The sum Snof the first ‘n’ terms of the GP is given by

Sn= a(rn-1)/ (r-1) , if r ≠1

= na, if r =1.
Harmonic progressionThe nth term of H.P. is given by the sequence a1, a2, …. an, where ai ≠ 0 for every i is said to be in harmonicprogression (H.P.) if the sequence 1/a1, 1/a2, …. ,1/an is in A.P.an=1 / [a+(n-1)d], where a = 1/a1and d = 1/a2-1/a1.Arithmetico geometric progression.
Suppose a1, a2, a3, …. is an A.P. and b1, b2, b3, …… is a G.P. Then the sequence a1b1, a2b2, …, anbnis said to be an arithmetic-geometric progression. An arithmetic-geometric progression is of the form ab, (a+d)br, (a + 2d)br2, (a + 3d)br3, ……

Its sum Snto n terms is given by

Sn= ab + (a+d)br + (a+2d)br2+……+ (a+(n–2)d)brn–2+ (a+(n–1)d)brn–1.

Multiply both sides by r, so that

rSn= abr+(a+d)br2+…+(a+(n–3)d)brn–2+(a+(n–2)d)brn–1+(a+(n–1)d)brn.

Subtracting we get

(1 – r)Sn= ab + dbr + dbr2+…+ dbrn–2+ dbrn–1– (a+(n–1)d)brn.

= ab + dbr(1–rn–1)/(1–r) (a+(n–1)d)brn

⇒ Sn= ab/1–r + dbr(1–rn–1)/(1–r)2 – (a+(n–1)d)brn/1–r.

If –1 < r < 1, the sum of the infinite number of terms of the progression is

limn→∞Sn= ab/1–r + dbr/(1–r)2.
```
6 years ago
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