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Arithmetic progression
For class 10.
In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant
Arithmetic progression - Introduction to Arithmetic progressions for class 10...Let's Tute
Arithmetic progression - Introduction to Arithmetic progressions for class 10 maths.
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Arithmetic progression
For class 10.
In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant
Arithmetic progression - Introduction to Arithmetic progressions for class 10...Let's Tute
Arithmetic progression - Introduction to Arithmetic progressions for class 10 maths.
Lets tute is an online learning centre. We provide quality education for all learners and 24/7 academic guidance through E-tutoring.
Our Mission- Our aspiration is to be a renowned unpaid school on Web-World.
Contact Us -
Website - www.letstute.com
YouTube - www.youtube.com/letstute
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Arithmetic progression
For class 10.
In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant
PROJECT (PPT) ON PAIR OF LINEAR EQUATIONS IN TWO VARIABLES - CLASS 10mayank78610
THIS A PROJECT BEING MADE BY INFORMATION COLLECTED FROM CLASS 10 MATHS NCERT BOOK.
THANK YOU FOR SEEING MY PROJECT ... I THINK THIS MIGHT HELP YOU IN YOUR HOLIDAY HOMEWORK PROJECTS .
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Presentation on Trigonometry. A topic for class 10 Students. Has every topic covered for students wanting to make a presentation on Trigonometry. Hope this will help you...........
Arithmetic progression
For class 10.
In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant
Probability - Question Bank for Class/Grade 10 maths.Let's Tute
Probability - Question Bank for Class/Grade 10 maths.
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SEQUENCE AND SERIES
SEQUENCE
Is a set of numbers written in a definite order such that there is a rule by which the terms are obtained. Or
Is a set of number with a simple pattern.
Example
1. A set of even numbers
• 2, 4, 6, 8, 10 ……
2. A set of odd numbers
• 1, 3, 5, 7, 9, 11….
Knowing the pattern the next number from the previous can be obtained.
Example
1. Find the next term from the sequence
• 2, 7, 12, 17, 22, 27, 32
The next term is 37.
2. Given the sequence
• 2, 4, 6, 8, 10, 12………
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2. Sequence: A list of numbers
having specific relation
between the consecutive
terms is generally called a
sequence.
e.g. 1, 3, 5, 7,……… (next term
to a term is obtained by
adding 2 with it)
& 2, 6, 18, 54,…….( next term
to a term is obtained by
multiplying 3 with it)
3. Arithmetic Progression: If various terms of a
sequence are formed by adding a fixed
number to the previous term or the
difference between two successive
terms is a fixed number, then the sequence
is called AP.
e.g.1) 2, 4, 6, 8, ……… the sequence of even
numbers is an example of AP
2) 5, 10, 15, 20, 25…..
In this each term is obtained by adding 5 to
the preceding term except first term.
5. The general form of an Arithmetic Progression
is
a , a +d , a + 2d , a + 3d ………………, a + (n-
1)d
Where ‘a’ is first term and
‘d’ is called common difference.
6. Common Difference - The fixed number which
is obtained by subtracting any term of AP from
its previous term.
If we take
First term of an AP as a
and Common Difference
as d,
Then,
nth term of that AP will be
An = a + (n-1)d
7. 3, 7, 11, 15, 19 …
Notice in this sequence that if we find the difference
between any term and the term before it we always get
4. 4 is then called the common difference and is
denoted with the letter d.
d =4
To get to the next term in the sequence we
would add 4 so a recursive formula for this
sequence is:
41 += −nn aa
The first term in the sequence would be a1
which is sometimes just written as a.
a =3
8. 3, 7, 11, 15, 19 …
+4 +4 +4 +4
Each time you want another term in the sequence you’d add d. This
would mean the second term was the first term plus d. The third term
is the first term plus d plus d (added twice). The fourth term is the first
term plus d plus d plus d (added three times). So you can see to get
the nth term we’d take the first term and add d (n - 1) times.
d =4
( )dnaan 1−+=
Try this to get the
5th term.
a =3
( ) 1916341535 =+=−+=a
9. Let’s see an example!!
Let a=2, d=2, n=12,find An
An=a+(n-1)d
=2+(12-1)2
=2+(11)2
=2+22
Therefore, An=24
Hence solved.
10. 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.
11. 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
12. The sum of n terms, we find as,
Sum = n X [(first term + last term) / 2]
Now last term will be = a + (n-1) d
Therefore,
Sum(Sn
) =n X [{a + a + (n-1) d } /2 ]
= n/2 [ 2a + (n+1)d]
13. DERIVATION
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)
Now let’s add (1) and (2):
2Sn
= [2a + (n – 1)d] + [2a + (n – 1)d] + …
……… + [2a + (n – 1)d]
So, S = n/2 [2a + (n – 1)d]
14. Problem . Find number of terms of
A.P. 100, 105, 110, 115,,………………
500Solution.
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
15. 5(n – 1) = 400
n – 1 = 400/5
n - 1 = 80
n = 80 + 1
n = 81
Hence the no. of terms are 81.
16. Problem . 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
= n/2 [ 2a + (n - 1)d ]
= ½ x 30 [ 2x 12 + (30-1)x 8]
= 15 [ 24 + 29 x8]
17. = 15[24 + 232]
= 15 x 246
= 3690
THE SUM OF TERMS IS 3690
18. Problem . Find the sum of terms in given A.P.
2 , 4 , 6 , 8 , ……………… 200
Solution: 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)
2(n – 1) = 198
n – 1 = 99, n = 100
19. The sum to n terms of an arithmetic progression
Sn
= n/2[ 2a + (n - 1)d ]
S100
= 100/2 [ 2x 2 + (100-1)x 2]
= 50 [ 4 + 198]
= 50[202]
= 10100
20. The difference between two terms of an
AP can be formulated as below:-
nth term – kth term
= t(n) – t(k)
= {a + (n-1)d} – { a + (k-1) d }
= a + nd – d – a – kd + d = nd – kd
Hence,
t(n) – t(k) = (n – k) d