# Arithmetic and geometric progressions *Sequences. Numerical sequences. General term of numerical sequence. * Arithmetic progression. Geometric progression. Infinitely decreasing geometric progression. Converting of repeating decimal to vulgar fraction.
*Sequences.* Lets consider the series of natural numbers: 1, 2, 3,
, *n* 1, *n *,
. If to replace each natural number * n * in this series by some number *u*_{n} , subordinated to some law, well receive a new series of numbers:
*u*_{1},* u*_{2},* u*_{3},
,* u *_{n }_{ - }_{1},* u *_{n} ,
,called a *numerical sequence*. A number *u*_{n} is called a* general term* of the numerical sequence. E x a m p l e s of numerical sequences: 2, 4, 6, 8, 10,
, 2*n*,
; 1, 4, 9, 16, 25,
, *n*² ,
; 1, 1/2, 1/3, 1/4, 1/5,
, 1/*n *,
.*Arithmetic progression. *The numerical sequence, in which each next term beginning from the second is equal to the previous term, added with the constant for this sequence number *d*, is called an * arithmetic progression*. The number *d* is called a *common difference. Any term of an arithmetic progression* is calculated by the formula:
*a*_{n} = a_{1}* + d ( n 1* )* . **A sum of n first terms of arithmetic progression* is calculated as:
E x a m p l e . Find a sum of the first 100 odd numbers. S o l u t i o n . Use the last formula. Here *a*_{1} = 1, *d* = 2 *.* So, we have: *Geometric progression. *The numerical sequence, in which each next term beginning from the second is equal to the previous term, multiplied by the constant for this sequence number *q*, is called a * geometric progression*. The number *q* is called a *common ratio. Any term of a geometric progression* is calculated by the formula:
*b*_{n} = b_{1}* q *^{n -}^{ 1}* .**A sum of n first terms of geometric progression* is calculated as:
*Infinitely decreasing geometric progression. * This is the geometric progression, with | *q* | < 1 . For it the notion of a *sum of infinitely decreasing geometric progression* is determined as a number, to which a sum of the first *n* terms of the considered progression unboundedly approximates at an unbounded increasing of number *n *. The infinitely decreasing geometric progression sum is calculated by the formula:
E x a m p l e . Find the sum of the infinitely decreasing geometric progression: S o l u t i o n . Use the last formula. Here *b*_{1}= 1, *q* = 1/2. So, we have: *Converting of a repeating decimal to a vulgar fraction. *Assume, that we want to convert the repeating decimal 0.(3) to a vulgar fraction. Consider this decimal in the more natural form:
This is the infinitely decreasing geometric progression with the first term 3/10 and a common ratio *q* = 1/10. According to the above shown formula the last sum is equal to: Back |