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GOBI ARTS COLLEGE 1994
Bachelor of Science (B.Sc.)
Gobi Arts college 1996
Master of Science (M.Sc.)
Ganapathy, Coimbatore, India - 641006
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
15
Board
IGCSE, CBSE, State
Subjects taught
Science, Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
15
Board
IGCSE, CBSE, State
Subjects taught
Physics, Mathematics
Taught in School or College
Yes
Answered on 20/08/2019 Learn CBSE - Class 10/Mathematics/Arithmetic Progression/NCERT Solutions/Exercise 5.2
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Post a Lesson
This A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 3
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a20 = a + (20 − 1) d
= 253 + (19) (−5)
a20= 158
Therefore, 20th term from the last term is 158.
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Answers 2 
Comments Answered on 19/08/2019 Learn CBSE - Class 10/Mathematics/Pair of linear equations in two variable/NCERT Solutions/Exercise 3.7
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Let the capital of the first =x
capital of the second = y
According to first statement,
x+100=2(y-100)
⇒ x-2y=-300------(1)
according to second statement,
y+10=6(x-10)
6x-y=70--------(2)
solving (1) & (2)
(2)×2⇒12x-2y=140
x-2y=-300
(2)-(1)⇒ 11x=440
x=440/11=40
substitute x=40 in (1)
40-2y=-300
-2y=-300-40=-340
y=-340/(-2) =170
First capital=40
second capital=170
Like 0 Answers 4 Comments Answered on 19/08/2019 Learn CBSE - Class 10/Mathematics/Arithmetic Progression/NCERT Solutions/Exercise 5.3
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Given a=5 l=45 Sn=400
Sn= n/2(a+l)
400 =(n/2) (5+45)
400=50n/2=25n
n=400/25
=16 = number of terms
To find common difference 'd'
a16=45 ⇒5+15d=45
15d=40
d= 8/3
Like 0 Answers 2 Comments Answered on 23/07/2019 Learn CBSE - Class 10/Mathematics/Arithmetic Progression/NCERT Solutions/Exercise 5.2
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Given a11=a+10d=38→(1)
a16=a+15d=73→(2)
(2)-(1)⇒ 5d= 35⇒d= 7
Substitute d=7 in (1) we get
a+10*7 =38
⇒a=38-70=-32
a31=-32+30*7=-32+210
= 178
Like 0 Answers 2 Comments Answered on 22/07/2019 Learn CBSE - Class 10/Mathematics/Arithmetic Progression/NCERT Solutions/Exercise 5.2
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The first 3 digit number divisible by 7 is 105
Last 3 digit number divisible by 7 is 994
formula- =a+(n-1)d
Number of terms n = (l-a/d) +1
=(994-105/7) +1
=889/7 +1
=127+1
=128
Hence there are 128 three digit numbers divisible by 7.
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
15
Board
IGCSE, CBSE, State
Subjects taught
Science, Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
15
Board
IGCSE, CBSE, State
Subjects taught
Physics, Mathematics
Taught in School or College
Yes
Answered on 20/08/2019 Learn CBSE - Class 10/Mathematics/Arithmetic Progression/NCERT Solutions/Exercise 5.2
Ask a Question
Post a Lesson
This A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 3
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a20 = a + (20 − 1) d
= 253 + (19) (−5)
a20= 158
Therefore, 20th term from the last term is 158.
Like 0 Answers 2 Comments Answered on 19/08/2019 Learn CBSE - Class 10/Mathematics/Pair of linear equations in two variable/NCERT Solutions/Exercise 3.7
Ask a Question
Post a Lesson
Let the capital of the first =x
capital of the second = y
According to first statement,
x+100=2(y-100)
⇒ x-2y=-300------(1)
according to second statement,
y+10=6(x-10)
6x-y=70--------(2)
solving (1) & (2)
(2)×2⇒12x-2y=140
x-2y=-300
(2)-(1)⇒ 11x=440
x=440/11=40
substitute x=40 in (1)
40-2y=-300
-2y=-300-40=-340
y=-340/(-2) =170
First capital=40
second capital=170
Like 0 Answers 4 Comments Answered on 19/08/2019 Learn CBSE - Class 10/Mathematics/Arithmetic Progression/NCERT Solutions/Exercise 5.3
Ask a Question
Post a Lesson
Given a=5 l=45 Sn=400
Sn= n/2(a+l)
400 =(n/2) (5+45)
400=50n/2=25n
n=400/25
=16 = number of terms
To find common difference 'd'
a16=45 ⇒5+15d=45
15d=40
d= 8/3
Like 0 Answers 2 Comments Answered on 23/07/2019 Learn CBSE - Class 10/Mathematics/Arithmetic Progression/NCERT Solutions/Exercise 5.2
Ask a Question
Post a Lesson
Given a11=a+10d=38→(1)
a16=a+15d=73→(2)
(2)-(1)⇒ 5d= 35⇒d= 7
Substitute d=7 in (1) we get
a+10*7 =38
⇒a=38-70=-32
a31=-32+30*7=-32+210
= 178
Like 0 Answers 2 Comments Answered on 22/07/2019 Learn CBSE - Class 10/Mathematics/Arithmetic Progression/NCERT Solutions/Exercise 5.2
Ask a Question
Post a Lesson
The first 3 digit number divisible by 7 is 105
Last 3 digit number divisible by 7 is 994
formula- =a+(n-1)d
Number of terms n = (l-a/d) +1
=(994-105/7) +1
=889/7 +1
=127+1
=128
Hence there are 128 three digit numbers divisible by 7.
Like 0 Answers 2 Comments Ask a Question
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