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'Lilavati' was one of the mathematical texts written by Bhaskaracharya-II (1114 - 1193 CE), a well-known Indian mathematician. In this presentation we have shared one of the techniques of multiplying numbers which has been explained in Lilavati.
To learn more techniques of Lilavati, join Vedic Mathematics Home Study Course. For details, visit vedicmaths.chinfo.org
Class 10 arithmetic_progression_cbse_test_paper-2dinesh reddy
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'Lilavati' was one of the mathematical texts written by Bhaskaracharya-II (1114 - 1193 CE), a well-known Indian mathematician. In this presentation we have shared one of the techniques of multiplying numbers which has been explained in Lilavati.
To learn more techniques of Lilavati, join Vedic Mathematics Home Study Course. For details, visit vedicmaths.chinfo.org
Class 10 arithmetic_progression_cbse_test_paper-2dinesh reddy
arithmetic progression class 10 cbse question paper for practice and is more easy to solve can gain extra talent in the subject maths especially in arithmetic progressions read this and be happy.
Worried about due completion of math homework? You can count on us. In Homework1.com we have excellent infrastructure to serve you even at the most critical hours of assignment submission! Try our service today and get excellent score in math exam.
For more instructional resources, CLICK me here!
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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………
Part 1 sequence and arithmetic progressionSatish Pandit
In this presentation, you will learn sequences and arithmetic progression. How to find the terms, common differences, etc. I have given detailed solutions to each problem.
This PPT will clarify your all doubts in Arithmetic Progression.
Please download this PPT and if any doubt according to this PPT, please comment , then i will try to solve your problem.
Thank you :)
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X std mathematics - Relations and functions (Ex 1.4), Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, composition of function, definition of function, composition of three functions, identifying the graphs of linear, quadratic, cubic and reciprocal functions, linear function, modules or absolute valued function, quadratic function, cubic function, reciprocal function, constant function
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This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
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http://sandymillin.wordpress.com/iateflwebinar2024
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Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
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2e. Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.5)
1. PEDAGOGY OF
MATHEMATICS – PART II
BY
Dr. I. UMA MAHESWARI
Principal
Peniel Rural College of Education,Vemparali,
Dindigul District
iuma_maheswari@yahoo.co.in
22. Solution:
To prove it is an A.P, we have to
show d = t2 – t1 = t3 – t2.
(i) a – 3, a – 5, a – 7………
t1,t2,t3
d = t2 – t1 = a – 5 – (a – 3) = a – 5 – a + 3 = -2
∴ d = -2 ∴ It is an A.P.
d = t3 – t2 = a – 7 – (a – 5) = a – 7 – a + 5 = -2
32. Answer:
First term (a) = 16
Common difference (d) = 11 – 16 = -5
tn = – 54
a + (n – 1) d = -54
16 + (n – 1) (-5) = -54
54 + 21 = -54
54 + 21 = 5n
75 = 5n
n = 75/5 = 15
The 15th term is – 54
33.
34. Answer:
tn = a + (n – 1)d
9 times 9th term = 15 times 15th term
9t9 = 15 t15
9[a + 8d] = 15[a + 14d]
9a + 72d = 15a + 210d
9a – 15a + 72 d – 210 d = 0
-6a – 138 d = 0
6a + 138 d = 0
6 [a + 23 d] = 0
6 [a + (24 – 1)d] = 0
6 t24 = 0
∴ Six times 24th terms is 0
35. Solution:
3 + k, 18 – k, 5k + 1 are in A.P
⇒ 2b = a + c if a, b, c are inA.P
36. Answer:
x, 10, y, 24, z are in A.P
t2 – t1 = 10 – x
d = 10 – x …..(1)
t3 – t2 = y – 10
d = y – 10 ……(2)
t4 – t3 = 24 – y
d = 24 – y …..(3)
t5 – t4 = z – 24
d = z – 24 …..(4)
37. From (2) and (3) we get
y – 10 = 24 – y
2y = 24 + 10
2y = 34
y = 17
From (1) and (2) we get
10 – x = y – 10
– x – y = -10 -10
-x -y = -20
x + y = 20
x + 17 = 20(y = 17)
38. x = 20 – 17 = 3
From (1) and (4) we get
z – 24 = 10 – x
z – 24 = 10 – 3 (x = 3)
z – 24 = 7
z = 7 + 24
z = 31
The value of x = 3, y = 17 and z = 31
39. Solution:
t1 = a = 20
t2 = a + 2 = 22
t3 = a + 2 + 2 = 24 ⇒ d = 2
∴There are 30 rows.
t30 = a + 29d
= 20 + 29 × 2
= 20 + 58
= 78
∴There will be 78 seats in the last row
40. When a = 9, d = 7
a + d = 9 + 7 = 16
a = 9
a – d = 9 – 7 = 2
When a = 9, d = -7
a + d = 9 – 7 = 2
a = 9
a – d = 9 – (-7) = 9 + 7 = 16
The three terms are 2, 9, 16 (or) 16, 9, 2
41.
42.
43. Solution:
Let the five days temperature be (a – d), a, a + d, a + 2d, a + 3d.
The three days sum = a – d + a + a + d = 0
⇒ 3a = 0 ⇒ a = 0. (given)
a + d + a + 2d + a + 3d = 18
3a + 6d = 18
3(0) + 6 d = 18
6d = 18
d = 18/6 = 3
∴The temperature of each five days is a – d, a, a + d, a + 2d, a + 3d
0 – 3, 0, 0 + 3, 0 + 2(3), 0 + 3(3) = -3°C, 0°C, 3°C, 6°C, 9°C
45. We find that the yearly savings is in A.P with a1 = 2000 and d = 600.
We are required to find how many years are required to save 20,000 a year …………..
an = 20,000
an = a + (n – 1)d
20000 = 2000 + (n – 1)600
(n – 1)600 = 18000
n – 1 = 18000/600 = 30
n = 31 years