Slide 1: Arithmatic Progression Class : X Presented by : Munish Sharma
TGT-Maths Name Of topic
Slide 2: An Arithmatic Progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except first term. Arithmatic Progression For example : 5, 10, 15, 20, 25…..
In this each term is obtained by adding 5 to the preceding term except first term
Slide 3: The general form of an Arithmatic Progression is
a , a +d , a + 2d , a + 3d ………………, a + (n-1)d Where ‘a’ is first term and
‘d’ is called common difference. nth Term of an A.P. is denoted by An
An = a + (n-1)d
Slide 4: The first term = a1 =a +0 d = a + (1-1)d Let us consider an A.P. with first term ‘a’ and common difference ‘d’ ,then To find the nth Term of an A.P. The second term = a2 = a + d = a + (2-1)d The third term = a3 = a + 2d = a + (3-1)d The fourth term = a4 =a + 3d = a + (4-1)d -------------------------------------------
------------------------------------------- The nth term = an = a + (n-1)d
Slide 5: To check that a given term is in A.P. or not. 2, 6, 10, 14…. Here first term a = 2,
find differences in the next terms
a2-a1 = 6 – 2 = 4
a3-a2 = 10 –6 = 4
a4-a3 = 14 – 10 = 4
Since the differences are common.
Hence the given terms are in A.P.
Slide 6: Which of the following are in A.P. ?
If they are in A.P. find their first term and common difference
1, 3, 9, 27…..
a, 2a, 3a, 4a…
-6, -2, 2, 6….
12, 22, 32, 42 ……
Slide 7: Problem : Find the value of k for which the given series is in A.P. 4, k –1 , 12 Solution : Given A.P. is 4, k –1 , 12
If series is A.P. then the differences will be common
d1 = d1
a2 – a1 = a3 – a2
k – 1 – 4 = 12 – (k – 1)
k – 5 = 12 – k + 1
k + k = 12 + 1 + 5
2 k = 18 or k = 9
Slide 8: Problem 1. Find 10th term of A.P. 12, 18, 24, 30.. … Solution. Given A.P. is 12, 18, 24, 30.. First term is a = 12
Common difference is d = 18- 12 = 6 nth term is an = a + (n-1)d
Put n = 10, a10 = 12 + (10-1)6
= 12 + 9 x 6
= 12 + 54
a10 = 66
Slide 9: Problem 2. Find number of terms of A.P.
100, 105, 110, 115,,………………500 Solution. Given A.P. is 100, 105, 110, 115,………………500 First term is a = 100 , an = 500
Common difference is d = 105 -100 = 5 nth term is an = a + (n-1)d
500 = 100 + (n-1)5
500 - 100 = 5(n – 1)
400 = 5(n – 1)
5(n – 1) = 400
Slide 10: Problem 3. Find the number of all three digit numbers divisible by 8 Solution. Smallest three digit number divisible by 8 is 104
& Largest three digit number divisible
by 8 is 992
Therefore numbers obtained are
104, 112,120, 128, …………………992
Which are in A.P.
Now solve using A.P. and find how many are
they ?
Slide 11: 5(n – 1) = 400
n – 1 = 400/5
n - 1 = 80
n = 80 + 1
n = 81
Hence the no. of terms are 81.
Slide 12: Let us solve some problems based on Arithmatic Progression 1. Find 50th term of A.P. 5, 10, 15, 20… 2. Find first three terms of an A.P. whose
3rd term is 10 and 6th term is 19. 3. Find number of terms of an A.P. 3, 6, 9,
………………..99.
Slide 13: Sum of n terms of an Arithmetic Progression Its formula is
Sn = ½ n [ 2a + (n - 1)d ] It can also be written as
Sn = ½ n [ a + an ]
Slide 14: You may need to be able to prove this formula. It is derived as follows:
The sum to n terms is given by:Sn = a + (a + d) + (a + 2d) + … + (a + (n – 1)d) (1)
If we write this out backwards, we get:Sn = (a + (n – 1)d) + (a + (n – 2)d) + … + a (2) The sum to n terms of an arithmetic progression
Sn = ½ n [ 2a + (n - 1)d ]
Slide 15: Now let’s add (1) and (2):
2Sn = [2a + (n – 1)d] + [2a + (n – 1)d] + … ……….. + [2a + (n – 1)d]
So Sn = ½ n [2a + (n – 1)d]
Slide 16: Problem 1. Find the sum of 30 terms of given A.P.
12 + 20 + 28 + 36……… Solution : Given A.P. is 12 , 20, 28 , 36 Its first term is a = 12
Common difference is d = 20 – 12 = 8 The sum to n terms of an arithmetic progression
Sn = ½ n [ 2a + (n - 1)d ]
= ½ x 30 [ 2x 12 + (30-1)x 8]
= 15 [ 24 + 29 x8]
= 15[24 + 232]
= 15 x 246
= 3690
Slide 17: Problem 2. Find the sum given A.P.
2 + 4 + 6 + 8 + ……………… + 200 Solution : Given A.P. is 2 , 4, 6 , 8 …………….200 Its first term is a = 2
Common difference is d = 4 – 2 = 2 nth term is an = a + (n-1)d
200 = 2 + (n-1)2
200 - 2 = 2(n – 1)
198 = 2(n – 1)
2(n – 1) = 198
Slide 18: The sum to n terms of an arithmetic progression
Sn = ½ n [ 2a + (n - 1)d ]
S100 = ½ x 100 [ 2x 2 + (100-1)x 2]
= 50 [ 4 + 198]
= 50[202]
= 10100 2(n – 1) = 198
n – 1 = 198/2
n – 1 = 99
n = 99 + 1
n = 100
Slide 19: Let us solve some problems based on sum of terms of Arithmatic Progression 1. Find sum of 40 term of A.P. 5, 10, 15, 20… 3. Find the sum of series 10 + 20 + 30 +… …………+ 1000 2. Find sum of 100 term of A.P. -70, -67, -64… 4. Find the sum of series -5 + (-15)+(-25+… …………+ (-525)
Slide 20: End Of Show Presented By Munish Sharma