IIT JEE MATHS
Open in Telegram
This Channel is for JEE(Main) / (Advanced) aspirants. You will get eBooks, previous year question papers, solutions, notes, etc. @jeeadvanced @jee_mains @jee_physics @jee_chemistry @jee_maths @jeequestions @jeequest @jeehindi @jee_quiz @studypandit
Show moreIndia117 927The category is not specified
5 375
Subscribers
No data24 hours
No data7 days
No data30 days
Posts Archive
5 375
NTA Abhyas JEE Main Test Question 2
Solution : https://youtu.be/2hJNsoMIZI8?t=120s
#Mathematics
@jee_mains
5 375
NTA Abhyas JEE Main Test 1 Question 1
Solution : https://youtu.be/2hJNsoMIZI8
#Mathematics
@jee_mains
5 375
Revision Notes on Arithmetic Progression
If βaβ is the first term and βdβ is the common difference of the arithmetic progression, then its nth term is given by an = a+(n-1)d
The sum, Sn of the first βnβ terms of the A.P. is given by Sn = n/2 [2a + (n-1)d]
If Sn is the sum of n terms of an A.P. whose first term is βaβ and last term is βlβ,Sn = (n/2)(a + l)
If common difference is d, number of terms n and the last term l, then Sn = (n/2)[2l-(n -1)d]
If a fixed number is added or subtracted from each term of an A.P., then the resulting sequence is also an A.P. and it has the same common difference as that of the original A.P.
If each term of A.P is multiplied by some constant or divided by a non-zero fixed constant, the resulting sequence is an A.P. again.
If a1, a2, a3, β¦, an andb1, b2, b3, β¦, bn, are in A.P. then a1+b1, a2+b2, a3+b3, β¦β¦, an+bn and a1βb1, a2βb2, a3βb3, β¦β¦, anβbn will also be in A.P.
Suppose a1, a2, a3, β¦β¦,an are in A.P. then an, anβ1, β¦β¦, a3, a2, a1 will also be in A.P.
If nth term of a series is tn = An + B, then the series is in A.P.
If a1, a2, a3, β¦β¦, an are in A.P., then a1 + an = a2 + anβ1 = a3 + anβ2 = β¦β¦ and so on.
In order to assume three terms in A.P. whose sum is given, they should be assumed as a-d, a, a+d.
Four terms of the A.P. whose sum is given should be assumed as a-3d, a-d, a+d, a+3d
Five convenient numbers in A.P. aβ2b, aβb, a, a+b, a+2 b.
In general, we take a β rd, a β (r β 1)d, β¦., a β d, a, a + rd in case we have to take (2r + 1) terms in an A.P.
Likewise, any 2r terms of an A.P. should be assumed as: a β (2r-1)d, a β (2r β 3)d, β¦., a β d, a, a + d, β¦β¦β¦β¦.. , a+(2r-3)d, a + (2r-1)d.
The arithmetic mean of two numbers βaβ and βbβ is (a+b)/2.
The terms A1, A2, β¦.. , An are said to be arithmetic means between a and b if a, A1, A2, β¦.. , An, bis an A.P.
Clearly, βaβ is the first term, βbβ is the (n+2)th term and βdβ is the common difference. Then, we have b = a+(n+2-1)d = a+(n+1)d
Hence, this gives βdβ = (b-a)/(n+1)
@jee_mains
