A 9th,10th class student can solve this JEE Mains 2023 Series | Aman Sir
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- Äas pĆidĂĄn 6. 02. 2023
- In today's video, we will be solving a Series Problem.
these Series questions was asked in JEE Mains 2023.
Even the 9th, and 10th class students can solve this problem.
How will you solve this Series Problem?
Give it a thought and apply concepts that you learned to solve Series Problem.
If you are unable to solve it, let's check out how Aman sir will make this Series Problem very easy to solve.
Check out the complete video to know the solution to the Inequality Problem.
A 9th,10th class student can solve this JEE Mains 2023 Series
Subscribe to @BHANNATMATHS for brain-twisting videos around maths.
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10th student I am and solving it made my day
10th student here.
Yes sir its very easy if we observe carefully.
You will study Sigma notations in 11th it will make you sum any series possible
Waiting for 21 March to start class 11 by our loving Aman sir â„ïžâ„ïžâ„ïžđȘđȘđȘ
Yes brother I also
Main bhi
Me too đ
àŠàŠźàŠżàŠ
@@gautamchoubey1275 try Mohit tyagi
Great video
Thanks a lotâ€.
Sir I am class 9th(school) and 10th(in coaching) couldn't solve this problem because i see first time this question.
Koi nahi
add the last three terms and the equation of series 3n-3 where 3 is d and a = 0 and and sum ofn will be 3n/2(n-1)
Toooo Easy
Thanks sir!!!
You didnt told that what to find in the question otherwise it was the only and only question of your channel that I would have solved
Sir I used different method:
AP: 1+2-3+4+5-6+............+(3n-2)+(3n-1)-3n
In these AP,
First, I have done the subtraction .i.e. 2-3, 5-6, 3n-1-3n.
Then I have add the sum .i.e. 1+(-1), 4+(-3), 3n-2+(-1).
Now,
New AP: 0+3+..........+3n-3
BY USING SUM FORMULA,
Sn = n/2 [2a+(n-1)d]
I got,
Sn = 3n/2 [n-1] or Sn = 3/2 (n^2-n)
Amar sir ke samjhaane ke tarika dekhkar pata chalata hai ki aman sir maith me bahut deep me chale gaye hai.main class 12 me nhi hu naahi jee kaa taiyari karata hu but tab par bhi padhata hu budhhi open ho jaata hau
Sir maine positive terms and negative terms ko alag kiya aur notice Kiya ki negative no.are multiples of- 3.So maine position ko+ 3 ka multiple ka sath add karwa ya and then -2(multiples of -3) apply Kiya.Then I took -3common from the negative terms and then the solution was like (1+2+3+5+5+6+7+8+9+.....)-6(1+2+3+4+5+6+7+8+9+.....).then maine add kiya
25 jan shift 1
It was easiest question in that shift in maths and was among few questions which were doable in 1 minute
Sir I solved by: The +ve are getting added in multiple of 3 and -ve no.s are also multiple of 3. Take 3 common we r left with (-2-3-4-5-6.....-2n-1+3n) which adds upto 3n(n-1)/2
sir sabse simple method 3 seperate ap bn jaaenge agar nth terms ki aur simply summation series lagana hai 3 line mein hojaaega question
Sir I solved it by using Summation technique
All thanks to MSM baba đ„đ„đ„đ„đ„
Arjuna , msm baba â€
Sir bring more questions for class 10â„ïžâ„ïž
Sir is solved it đ
awesome method sir m to 3 bar sigma k prayog s nikal rha tha faltu m..thankyou so muchđ
đ„
I am in 10th couldn't solve it but i will try harder next time
Sirji hit and trail se bhi ho jayega.
n ko 2 put karenge toh answer aa jaayega.
Add and subtract 2 Ă ( 3+6+9...+3n)
Then it will become
3n(3n+1)/2 - [3n(n+1)/2]Ă2
= 3n/2(n-1)
sir n = 1 daalne par yeh pheli 3 terms ka sum de rha hai similarly n =2 dalne pr first 6 terms ka sum de rha hai , esa to nhi hona chahiye sir
What if we use Sigma notations instead
I have done it by summation concept
sir we can"t we also do this question by breaking it into three series
first=1+4+7.......+3n-2
second=2+5+8........+3n-1
third=-3 -6 -9............-3n
Sir I got 97.98 Percentile only because of youâ€
99.97 here đ
@@Ok....- how much marks
@@harshitrepswal 284
@@Ok....- wtf
@@rajistired yes
â€ïž
Good teaching
My You tube Channel MATHMATICS UNTOLD
i think this can be just simple solved by the following method :-
S = given series in the Q
s1 = 3+6+9+12 +15 .... +3n
subtract and add 2 times of S1 from S
we just get two simple series in ap as follows :-
(1+2+3+4+....3n) - 2(3+6+9+12...+3n)
i have solved the problem by this method and i have also got the same answer
Bro, in this series of :
1+4+7...(3n-2); how to get the total no. Of terms so that we can put it in the formula to get the sum
@@dhonikumarshahi2806 this is an AP with starting term = 1
And with common difference = 3
Mth term of ap will be :-
A(m) = 1+ (m-1).3
3n-2 = last term
3n-2 = 3m-2 ( simplified RHS of A(m))
n= m
So the number of terms is n ...
@Jay Thakkar yeah,
But did you perform this calculation or just applied any sort of logic to get n = m
@@dhonikumarshahi2806 tbh , this calculation happened in my unconscious mind( in seconds ) and i just showed that .... Not any other form of logical short cut
@Jay Thakkar Good,
I also thought the same but had to do all these calculations for the conformation.
Btw are you a jee aspirant ?
I am a class 10th student. Was able to do it easily
Ye konsi shift me aaya tha
Hello Bro...! I like your way of presentation. I am also a teacher. The smart board you are using is a nice one. I want to buy one such board. Would you please send me the link where I can buy it from?
Thanks and regards.
5 months and no response đđ€Łđ€Łđ€Ł
You can also find the value of n... This is possible as the sum of this series comes out to be 1/4 (by ramanujan paradox equation B [Available in Aman Sir's Ramanujan paradox video])...
So the value of n is (-1± â3)/2
n must be an integer..đ
N shouldn't be negative
@Zero Plays Are oye murkh baalak 'n' ki value kuch bhi kaise bol raha hai wo paradox hai konsa theorem nhi jo ki tu
use Karke n value nikal sakta hai
â@@kshitij7203 đđ
sir answer to mera bhi yehi aaya, but agar humare final function mai n=3 daalenge to 9 aa raha hai. but aise dekhe to 1+2-3 0 aana chahiye na?
Bro n=3 ke liye [(1+2-3)+(4+5-6)+(7+8-9)] ye add hongee.....ya phir Final series me n=3 ke liyee 0+3+6.. aur jaise n=4 ke liyee..{0+3+6+9}
sir 0 ka koi contribution nhi h toh usko hataya toh bhi chalega na.. 3 ko 1st term consider karna chahiye
Fir number of terms kum ho jayenga
Jaise 0 , 1 , 2 , 3 ,4 mai 5 terms hau toh n = 5 hoga
Ab zero hata diya jaye toh n = 4 lena padega
Toh iss question mai zero nahi hataya kyunki "n" lena jada aasan hai "n-1" lene se
I am 10th class student and solve this ap in first attempt , thank,s sir
This is wrong. How can we assume the sum is always going to end in a multiple of 3 terms?
3rd from formula is 9 while in the actual series written on board is 6. Why? Hope solution is not correct.
Sir jee advance ke liye series laeye plzzzzzz
Easy peasy
Sir I did it with another approach where
A= 1+2+3+4+5+.......3n
Let Sum of this series be X which is 3n(3n+1)/2
Then B = 3+6+9+12..... +3n
Let the sum of this series be Y which comes out to be 3n(n+1)/2
Now I did X-2Y
Which comes out to be 3n(n-1)/2
-class 10th
Bro 2 Y kyu kiya
@@khanitparashar2995 kyuki usne pahle y wali series jo actual me substract honi thi usko add kar diya to wo series actual series se 2y jyada hogyi thi
jaise ki 1+2-3=0
ek baar 3 add karoge to hoga 1+2=3
ek baar aur 3 add karoge to hoga 1+2+3=6
i hope ab samajh aya hoga
Easy hi question tha
@Sumit Shakya and @Dhanajay Sharma @kp -You can also find the value of n... This is possible as the sum of this series comes out to be 1/4 (by ramanujan paradox equation B [Available in Aman Sir's Ramanujan paradox video])...
So the value of n is (-1± â3)/2
@@zeroplays9915 but n ki value natural number honi chaiye na
Sir advance ki series lao
Easy sawaal
đ
I solve three different series to get the sum
à€źà€à€Ÿ à€ à€à€Żà€Ÿ à€čà„
Am in class 10 sir, why you taken a=0 instead of 3 in new A.P
if we take a=3 instead of a=0 then our total terms will be n-1 not n .
we have n terms including 0 as first term .
if we want to neglect 0 then totatl terems will be n-1
@@mr.nandu2607 ok thanks
Asan tha
Me be like :- Yeh question hai ?
Sir mera 72 percentile obc aaya h kya mai advance ke liye qualified hu?
Reservation ka chakkar Babu bhaiya đŹ
99.97 here đ
we can also do it like this:
(1+4+7...3n-2)+(2-3+5-6+...3n-1-3n)
in the first one, there are n terms and in second there are 2n terms
(n(3n-2+1)/2) + (-1-1-1-1...-1)
in the -1 series, there are 'n' number of -1.
n(3n-1)/2 + (-1 x n)
n(3n-1)/2 -n
(3n^2-n-2n)/2
(3n^2-3n)/2
3n(n-1)/2
I am also in 10th
Sir humare 9 class me ye chapter hi nhi hai
Icse ki bt krrhe wo
Class 10th CBSE ke portion mai hai, 9th ke nahi.
Allen me ye sb 6th me pdhate hai đ
I am a 9th student
Let's see if I can solve...
Sorry I wasn't
If you study only 9th then it's not solvable because at last an AP formula is used......You study it in 10th so don't worry if you failed to solve it....
However only a brilliant student of class 9 can solve this so if you solved it by just studying 9th then well done
Same here ye humare slyabus me hai hi nhi
Me to class 8th me padhata hun mujhe iska sol ata hai mere papa ne iye pucha tha he is maths teacher
@AkâŽâ·_shat But formula comes in 10th.....How come you studied it in 9th
Maybe you study ahead of your classđ
@AkâŽâ·_shat ooo....Its good to study extra as it helps later
I have an alternate method to do this...
Put 1 at the place of n in ans we get = 0
Now check ki wo question me baith rhi hai ki nahi...
This series not deserve a dedicated video
Sir mote ho rahe hai
Worst teacher
Kya hua brođ
You are worst đ„đ„đ„
Abe pagal h kya đ
Depressed ho kya bhai?