Maths

 Arithmetic progressions

INTRODUCTION:-

An arithmetic progression is a sequence in which terms increase or decrease regularly by the same constant. This constant is called the common difference of the retrogression (series).

In other words an arithmetic progression is the list of number in which the first term is given  and each

term is obtained by adding a fixed number d to the preceding term.

This fixed number d is called the common difference. This can be positive, negative, or zero.

EXAMPLE:-

1) 1,2.3,4,5......20

here d =1 and a (first term)=1

2) -1.0,-1.5, -2.0,-2.5

here d= -0.5 and a=-1.0

GENERAL FORM :-

a,a+d,a+2d, a+3d,a+4d,......

It represents an arithmetic progression where a is the first term and d is common difference . This is called the general form of an AP.

FORMULAE :-

1) d = a(2)-a(1)

2) a(2) = a + (2-1)d  , a(3) = a + (3-1)d

3) a(n) = a +( n- 1)d

To find  the sum:-

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

2) s= n/2[a +l ]

QUESTIONS:-

Q1) write first four terms of the AP , when the first term a and the common difference d are given as 

follows ?

1) a = 10 , d = 10

2) a = 4 ,  d = -3

Q2) which term of the AP : 3,8,13,18,...., is 78 ?

  

Q3)  200 logs are stacked in the following manner ; 20 logs in the bottom row , 19 in the next row 18 

in the row next to it and so on . In how many rows are the 200 logs can be arranged and how many logs are in the  top row? 

SOLUTION:-

sol1) we know that if the first term is a and the common difference is d. then a.a+d.a+2d .a+3d....

represent an AP for different values of a and d.

(1) putting a = 10 and d = 10 in a,a+d , a+2d , a+3d.....we get the required AP as 10,10+10,10+2*10...

i.e.,10.20.30,40....

(2)putting a = 4 and a = -3 in a,a+d,a+2d, a+3d.....we get the required AP as  -2,-2+0,-2+2*0, -2+3*0

i.e. -2,-2.-2.-2.....

sol2) we have: a = 3, d = 5

let 78 be the nth term of the given AP . Then,

a(n) = 78

therefore  3 + (n-1)5 = 78

                 5(n-1)= 75

                  n= 15+1

                  n= 16

Thus, 78 is the 16th  term of the given AP .

sol3) clearly,log stacked in each row from a sequence 20, 19,18,17,.....it is an AP with a= 20 and d = -1

let s(n) = 200. Then,

n/2[2*20+(n-1)(-1)]= 200

n^2-41n+400= 0

(n-16) (n-25) =0

i.e., n= 16 or 25

n = 25 is not valid for this problem 

therefore  no. of logs in the 16th row =a (16)

                                                             = a + 15d =20 + 15 (-1)

                                                             = 20-15=5