Progressions:
“A teacher
affects eternity
He can never tell when
his influence stops.”
-
- - Henry Adams
-
There are 3 progressions namely, Arithmetic
Progression, (AP), Geometric Progression (GP), Harmonic progression (HP).
Let
us discuss one by one. Before we define the AP, let me say what a sequence is? what
a series is, this would facilitate to understand better.
A sequence is an arrangement of numbers in a
particular order. A series is defined as the sum of the elements of a sequence.
Its applications are there in our daily life. It helps
us to analyse a pattern. Building a ladder with sloping sides – the
length/distance between two sticks /landing remains the same. Multiples of a
number say 5, are 5, 10, 15 ,20 - - - -, house numbers in a street, monthly rent,
increase of salary for every year, page number of a book, annual measurement of
the diameter and circumference of a tree trunk form an AP, foot prints of
animals / people along a track the distance covered be in AP, paying an
interest for every month if anyone has availed a loan. Travelling in a taxi one
has to pay initial rate plus rate per kilometer shall be added. It forms an AP;
waiting for a bus predicting when the next bus would come provided the traffic
moves with constant speed. We know in a clock; time progresses in perfect AP
second by second. Sum of AP series, its application, we do use in research in
science subject, nuclear science, space science.
In a sequence if the terms are such that the
difference between two consecutive terms is constant, (to be specific
difference between the second term and the first term, difference between the
third term and the second term, so on), we say that the terms are said to be in
AP. The difference is called as the common difference (cd) and is denoted by
“d’.
Eg: 4,7,10,13,
cd - - - - - - - - - - cd is 3;
2, -1, -4, -7 - - - - - - - - cd is -3 ( -1 -
2)
-3,1, 5,
9, - - - - - - - - - .cd is 1 –(-3) = 1 + 3 = 4,cd is 4.
In general, we form a sequence as, a, a+ d, a+2d, a+3d
,…., cd is d.
To find the nth term of an AP: we can proceed in this way.
Given the first
term as 6 and common difference is 3, fine the nth term.
The sequence is 6, 9,12, 15, -------------.
1 st term is 6,
2 nd term is 6 + 1(3) = 6 + 3 = 9
3 rd term is 6 + 2(3) = 6 + 6 = 12.
4th term 6 + 3(3) = 6 + 9 = 15., proceeding
in this way, to find
nth term 6 + (n-1) 3 6 + 3n -3 = 3n +3.
We can generalize as nth term is a +( n-1)
d - - - - - - - - - - I
There is a formula to find the sum to n terms of an
AP:
Let us take
AP as a, a+ d, a+2d, a+3d,
----------------------------------(1)
nth
term is a + (n-1) d, let us denote by ‘l’
ie we can write sum of sequence 1 as S= a+a+d+a+2d+a+3d - - -, l
(2)
Rewriting the sum of sequence 2 from end S = l+l-d+l-2d+l-3d, l-4d ------, a (3)
Adding (2) and (3) we get n times of (a+ l) = n (a+ l).
If S denotes the sum of series, then we get 2 S = n (a + l), hence sum to n terms is given as,
S = n/2 (a + l), substituting for l. we get S = n/2.{ 2a + (n-1) d}
Note: If a, b, c are three consecutive terms in AP,
then b = (a + c) / 2; b is called as arithmetic mean of a and c.
Problems for practice:
1. Find 40 th term
of the AP 8, 5, 2, - -,- - - - - -
- -
Cd is 5 – 8 = - 3, first term is 8, using the formula
we get,
40 th term = 8 + (40 -1) ( -3) = 8 + 39 x – 3 = 8 -
117 = - 109 is the answer.
2 Find the
value of n, given that - 49 is the n th term. AP is 11, 8, 5, - - -
- - -
Using the
formula again n th term, a + (n -1) d = - 49, here a = 11 and d = - 3.
On
substituting we get, 11 + (n -1) x (- 3) = - 49, 11 – 3n +3 = -49,
- 3n + 14 = - 49, on
simplification we get n = 21 is the answer (for - 3n = - 63)
3. If 3, x, y, 18 are in AP find
the values of x and y.
We can use the arithmetic mean
formula. x = (3 + y) / 2, y = (x + 18) / 2
On simplification we get 2 x = 3 + y ( i) , 2y = x + 18, substituting for y in
equation (i) we get 4 x = 6 + x + 18 , the value of x is 8 , hence value of y is 11
.
4.Is 280 a term of the AP
? 11+15+19+23+27 + - - - - - - - - - -
-
Ans: a = 11 and cd = 4. Let us find the
which term of the sequence we get 280?
To use n th term formula. 280 = 11 + (n-1) x (4); 280 = 11 + 4n -4= 4n
+7;
280 – 7 = 4n, 273 = 4n , n = 273 / 4 =
68 .25, n cannot be a fraction ,hence we
can
Conclude 280 is not term in the given
sequence.
5.Find the sum of all integers
between 50 and 100.
The given series is 51+52+53+54+55+ - - - - - - - +- 99 ------(i)
Here a = 51 d = 1, Let S denote the sum.
using the formula sum to n terms of an AP
S = (49/ 2) x {51+ 49} =
49 x 50 = 3675 is the answer.
6. Find the sum of the first 50 odd numbers.
Ans: The given sequence is 1 +3+5+7+9+11+13+ - - - - - - - - - +99.
There are 50 odd numbers, we can use the
formula (n/2) (a +l) l denoting last term.
S
= 50/ 2 (1 +99) = 25 x 100 = 2500 is the sum.
-
-
- - - - - - - - to be continued GP.
