I attempted as follows ...mentally By observing 4 n and 4 n -1 terms are positive 4n -3, 4n-2 terms are negative Hence summation of (4 n)^2 cancels now Σ(4n)^2 + Σ(4n-1)^2 - Σ(4n-2)^2 - Σ (4n-3)^2 becomes Σ (-8n+16n+24n) + Σ (1-4-9) =Σ (32n) - Σ(12) =16n(n+1) -12n S =4n(4n+1) Clearly putting n=8 S=32×33=1056. Option a and n=9 S= 36×37=1332. Option d Other values do not come
SIR btw for first question, the sum will be in form of -1^2-2^2+3^2+4^2 and so on if we add 1^2 + 2^2 + 5 ^2 + 6^2 and so on and subtract the same, we get Sn= 1^2+2^2+3^2+4^2+5^2 and so on till 4n terms -2(1^2+2^2 + 5^2 + 6^2 +... ) Sn= 4n(4n+1)(8n+1)/6-2(X) [ let the other sum be X] Sn=2[n(4n+1)(8n+1)-3X] / 3 this shows that since Sn is whole number, it must be even(since multiplied by 2) and must be a multiple of 3 . Since Sn is multiple of 3, we only find (a) and (d) options to be true.
I am writing this comment before watching the video to share that I solved this question correctly in my second attempt without seeing the solution and got correct answer.😁 Love these kind of videos and especially the confidence u get when you solve before watching the video. ❤
Okay Not gonna lie I solved in first try and Find my answer but at 6:17 I got heart attack as I thought I got wrong answer but After consulting many times I was sure I was Correct so sir kindly check this at there must be (4K+4)^2 = 16(K+1)^2 THnak you sir for this session
Sir series ke questions aate par analyse karke nikal liya yekadam perfect ans aaya AAP Kam time jo aspect/ jiska proof aspect hai o sab kuch 4-6hour me sikha dete ho AAP ko selute👮♂️👮♂️
I think 2nd que me solve jrte waqt while finding the common terms, hum ap's ko likhte hai, and ek ke n ko dusre ke m me likhte hai and we hit and trial for a natural no. Is sawal me jb hum aisa krne ki koshish krenge, toh, 2018 ko less than equal to rkhenge coz, n ki value jo bhi ho wo 2018se zyada nhi hoskta. n ki value find krenge toh ek me 2018 se zyada value ayegi. Usko reject krna tha. And dusre se krna hoga phir. I think exam ke time pe pressure hota hai and usme ye sb sochna and agar zroort pade toh approach ko badalna thoda mushkil hoskta hai.
Solution to the problem. n(x U y) = n(x) + n(y) - n(x ∩ y) n(x) + n(y) = 4036 We have to find the number of terms in AP1 = AP2 which mean 1 + ( k1 - 1 ) * 5 = 9 + ( k2 - 1 ) * 7 , where k1 and k2 are k1_th and k2_th term of AP1 and AP2 respectively. On solving it, 5k1 = 7k2 + 6. putting some initital values of k1 and k2 which satisfies 5k1 = 7k2 + 6, we get the below table k1 k2 4. 2 11. 7 18 12 25. 17 ... and so on. observe that both k1 and k2 are in AP. let's consider k1 for now, which has first term = 4 , common difference = 7. now we want to know the number of terms with the sequence 4 , 11 , 18 , 25 ... 4 + ( x - 1 ) * 7 x maximum intersection terms = 288 now final answer = 4036 - 288 = 3748
Sir mene terms ko rearrange Kiya like 3^2 - 1^2 + 4^2 - 2^2.... Fir (3-1)(3+1) + (4-2)(4+2)... 2(3+1) + 2(4+2)...... 2(1+2+3+4....) 2×n×(n+1)/2 That is n(n+1) Fir n ke value putting karke check kiya
In the first question, if you add the first 4 terms and the next 4 terms you will get a new series whose first term is 20 and common diff 32 upto n terms hence Sn=n(4+16n) now put n=8 and 9 therefor ans is option A and D
at 7:30 on screen simplification nahi kiya kyuki hua hi nahi hoga, sir ne 4(k+1)^2 likha tha, vo actually (4k+4)^2 tha = 16(k+1)^2 frame change me 2 ans bhi mark kar diye
Sir your teaching technique is very good and easy to understand. I am in 11 and solved the last question. a=16 d=5×7=35 for x tn=10086 and for y tn=14128 . So the ap formed with first term 16 and common difference 35 will consist of last term less than 288.71. So terms of XnY is 288. Then terms of XuY is 2018+2018- 288 as you directed. The ans comes out is 3748.
@@aadityamishra4633 sets, Binomial theorem, complex no, sequence n series, limits, 3d geometry, differentiation, trigonometry, quadratic eq etc. in offline. But at jee level , I completed 6-7 chapters.
Sorry but I have a doubt You put 4²+8²+......+(4n)² as 4(k+1)² [0,(n-1)] But when I substitute the value of k in the equation like when k=0 => 4(k+1)²=4 but shouldn't it be 4² Similarly, when k=1 => 4(k+1)²=16=4² but shouldn't it be 8² ? 🤔🤔🤔 Well I'm confused if anyone know what I am saying wrong or if there is a mistake in sir answer please explain. Thanks 😊
Bhai 1st time me too nahi bana But resume karne ke baad aapka pehela step dekha Uske baad too apne aap kar liya 🥲🥲🥲 But ye samaj nahi aa raha Ye 2nd wala quesⁿ itne bacchoan ka galat kese ho sakta hai Esse easy easy quesⁿs too Mere DPP level - 1 ki starting me hee hote hai 😅😅😅 JEE aspirant 2025 🔥🔥🔥
The question is not actually tough but the main step was a bit intuitive. If you get that step then your question is done. And I actually got this question in my first try. Feeling confident. By the way thank you bhaiyya for the solution.
Sir maine puri series ko 2 halves me toda aur dono ka sum krke Sn nikala to vo product of two consecutive no. aya 2n(2n+1) fr hit and trial lagaya thoda dimaag laga ke and answer comes out to be 37*36=1332 aya pr pahle pr nhi dhyan diya🤧
agr is question me AP nhi bnti dikhe Tb bhi sigma ko negative sign pr break kr k Dono side ki general terms likh k subtract kr skte isse bhi sum easily aa jati h or phir bs quadratic me value rkhni h
i did it in a completely wrong way but i got the answers in 30 secs💀💀 like i Hadnt considered that negative i oonly saw k^2 and you know that sum of it will be divisible by 6 and when i checked the answers i found A ,D correct but i thought it is wrong bcoz i hadnt considered negative complex term but i am hella completly right
Hello in the first question cant we do by grouping? Like taking 1st and third digit together and applying a^2-b^2 so we get all 2 as common and then its just normal sumation of 1 2 3 4 5...sir please correct me if I am wrong...although I am not getting desired answer from this but my method seems legit
are bhai mera bhi 288 par yeh galat tha kyukin x union y pucha hai = 2018+2018-288 x intersection y = 288 nahi silly mistake se hi sabka galat hua HAI !!!!!!
Sir first wala ques to bhot halwa tha agr aap thoda sa or ache se observe kar lete to 😂😂 Dekho .. (-1²-2²+3²+4²)+(-5²-6²+7²+8²).... Matching 1st term with third and and 2nd with 4th and so on .... applying a²-b²=(a-b)(a+b)... =>(3-1)(3+1)+(4-2)(4+2)+(7-5)(7+5)+(8-6)(8+6)......(4n)².. Taking 2 commin from each term =2(3+1+4+2+7+5+8+6.....)in this sum series every natural number came till 4n... Applying Sigma 2sigma(sum of natural numbers till 4n)=2(4n)(4n+1)/2=... Pitting n=9,8,10 will give us required answer.. I think my aprroqch is far better than you sir but I don't claim it my teacher solved it last year in class with us ..I gave due respect tou you sir but I was just giving a better approach for this ques for your students and you too Love you sir.
@@SudiptaDhar-q5n Never doubt on Msm sir( Baba)from pw🤐 Now sankalp Bharat... Best teacher I have seen in my life for maths ..jai baba ki ..jbk jbk.. Agr aapko kuch lga ho ki galat h to pls inform Kari agr hoga to I will ensure it must be corrected else you should see where you find the mistake ,may be your concept is losse or there might be some other reasons that you said that my solution bus wrong...
Sir answer 288 (edit-(sorry 2*2018-288) maine intersection le liya tha!!)hoga by observing the equation 5n1-7n2=6 Note- I got this eq by equating the n1th term of AP1 and n2th term of AP2!! SO/-Isme 5n1 keval 0 aur 5 Unit digits vale no lega isliye 7 ke multiple me ham keval 2 and 7 vale no dekhenge (n2) and by putting values for n2 we get an AP series with First no = 4 and common diff =7 to usse maine 2018 ki position pata ki to vo 288.something aayi to syd yhi answer hona chahiye tq sir🥱 SIR PLEASE PIN THE COMMET IF THE MY OBSERVATION IS RIGHT!! 👍
In the first question, i identified the pattern but still wasn't able to do it, as i didn't wrote it the way u did and hence wasn't able to notice that there will be 4 sequences And in the 2nd one, i got the answer, it was pretty simple ngl, and I'm wondering why so many people weren't able to do it
Bhai mai class 10th maharashtra stateboard se padha hu aur ab directly maths karne jara hu jee level ki to nahi hori hai boht demotivation hora hain please start 11th maths in detail on this channel as I LOVE THE WAY U TEACH🙏🙏🙏🙏🙏🙏
😂😂I solved it in a very much easier way.. first me jo series banai h.. uska last, second last and third last terms bhi likhlo by observing pattern. fir series 2nd,3rd and 1st ,4th ko a2 - b2 ka formula lagao yahi pattern follow karo pure series me se 1+3 common aa jayega and bracket me ek nayi ap banegi of 5+13+21...n terms. Solve karne pe 4n(1+4n) answer aata h..
me after doing advance calculus integrals involving the gamma, beta, diagama function.. : seprate into sums of odd and even, you just need to check the convergence criteria for which the series converge.. lmao technical talks
I attempted as follows ...mentally
By observing
4 n and 4 n -1 terms are positive
4n -3, 4n-2 terms are negative
Hence summation of (4 n)^2 cancels now
Σ(4n)^2 + Σ(4n-1)^2 - Σ(4n-2)^2 - Σ (4n-3)^2
becomes
Σ (-8n+16n+24n) + Σ (1-4-9)
=Σ (32n) - Σ(12)
=16n(n+1) -12n
S =4n(4n+1)
Clearly putting
n=8 S=32×33=1056. Option a
and
n=9 S= 36×37=1332. Option d
Other values do not come
same method but not mentally..on paper..i entered 11 th class 1 month back
SIR btw for first question,
the sum will be in form of -1^2-2^2+3^2+4^2 and so on
if we add 1^2 + 2^2 + 5 ^2 + 6^2 and so on and subtract the same, we get
Sn= 1^2+2^2+3^2+4^2+5^2 and so on till 4n terms -2(1^2+2^2 + 5^2 + 6^2 +... )
Sn= 4n(4n+1)(8n+1)/6-2(X) [ let the other sum be X]
Sn=2[n(4n+1)(8n+1)-3X] / 3
this shows that since Sn is whole number, it must be even(since multiplied by 2) and must be a multiple of 3 . Since Sn is multiple of 3, we only find (a) and (d) options to be true.
Ans- First common term is16 common difference of common ap is LCM{5,7} so Tn1=10086,Tn2 =14119 Tn1
3748 bhai answer
@@telugumyths5730 XnY = 288 hoga aur ans will be 3748
@@telugumyths5730 ha ha thoda mistake hogaya
Easy
X intersection Y is 288 and answer is 3748
Small correction: 4th term should be (4k+4)^2 = 16(k+1)^2
Not 4(k+1)^2
I also noticed bro
I am writing this comment before watching the video to share that I solved this question correctly in my second attempt without seeing the solution and got correct answer.😁
Love these kind of videos and especially the confidence u get when you solve before watching the video. ❤
Okay Not gonna lie I solved in first try and Find my answer but at 6:17 I got heart attack as I thought I got wrong answer but After consulting many times I was sure I was Correct so sir kindly check this at there must be (4K+4)^2 = 16(K+1)^2
THnak you sir for this session
Exactly i went for this otherwise the first term is 4 and according to Sn there is 16
You observed it🎉
you got me bruhhh!!! i got the same thought during solving.....
Yeah
Yeah I noticed it too👍
-12k^2. +8k + 8 solve karke yeh a rha..... 💀
Equations ko...
Sir series ke questions aate par analyse karke nikal liya yekadam perfect ans aaya
AAP Kam time jo aspect/ jiska proof aspect hai o sab kuch 4-6hour me sikha dete ho
AAP ko selute👮♂️👮♂️
I think
2nd que me solve jrte waqt while finding the common terms, hum ap's ko likhte hai, and ek ke n ko dusre ke m me likhte hai and we hit and trial for a natural no.
Is sawal me jb hum aisa krne ki koshish krenge, toh, 2018 ko less than equal to rkhenge coz, n ki value jo bhi ho wo 2018se zyada nhi hoskta. n ki value find krenge toh ek me 2018 se zyada value ayegi. Usko reject krna tha. And dusre se krna hoga phir.
I think exam ke time pe pressure hota hai and usme ye sb sochna and agar zroort pade toh approach ko badalna thoda mushkil hoskta hai.
S1= 1,6,11,16,21,26,31,36,41,46,51 upto 10086
S2= 9,16,23,30,37,44,51 upto 14128
Common AP = 16, 51 (a=16 , d=35)
since the common term can only be upto 10086 so,
a+(n-1)d < 10086
16 + (n -1) 35 < 10086
n-1 < 10070/35
n-1 < 287.7
n < 288.7 (288)
hence (S1 ∩ S2) = 288
(S1 U S2) = S1 + S2 - (S1 ∩ S2)
(S1 U S2) = 4036 - 288
(S1 U S2) = 3748
Thank you 🙌🙌
9:54 The answer is 3748 not 288 as X U Y is 2018*2 - ( common terms i.e. 288)
No
Answer is 3749
Solution to the problem.
n(x U y) = n(x) + n(y) - n(x ∩ y)
n(x) + n(y) = 4036
We have to find the number of terms in AP1 = AP2 which mean 1 + ( k1 - 1 ) * 5 = 9 + ( k2 - 1 ) * 7 , where k1 and k2 are k1_th and k2_th term of AP1 and AP2 respectively.
On solving it,
5k1 = 7k2 + 6.
putting some initital values of k1 and k2 which satisfies 5k1 = 7k2 + 6, we get the below table
k1 k2
4. 2
11. 7
18 12
25. 17
... and so on.
observe that both k1 and k2 are in AP. let's consider k1 for now, which has first term = 4 , common difference = 7. now we want to know the number of terms with the sequence 4 , 11 , 18 , 25 ... 4 + ( x - 1 ) * 7 x maximum intersection terms = 288
now final answer = 4036 - 288 = 3748
Mera bhai and also best teacher (for me) for a reason 🗿🗿
Using chinese remainder theorem the common numbers are of the form 35k+16 , which is atmost 10086 , kmax = 287 total 288 numbers
Same
Yeaah this is the Olympiad mathematics room right. I did it the same way. CRT is OP
I thought all of them were quite straightforward and doable in 4 mins each
Could you please elaborate this theorem
@@sourthakjackmcober29from where did you learn that bro
it was my class illustrations..thank you bhuwnesh sen(BS Sir)
Sir mene terms ko rearrange Kiya like 3^2 - 1^2 + 4^2 - 2^2....
Fir (3-1)(3+1) + (4-2)(4+2)...
2(3+1) + 2(4+2)......
2(1+2+3+4....)
2×n×(n+1)/2
That is n(n+1)
Fir n ke value putting karke check kiya
In the first question, if you add the first 4 terms and the next 4 terms you will get a new series whose first term is 20 and common diff 32 upto n terms hence Sn=n(4+16n) now put n=8 and 9 therefor ans is option A and D
Did it same way bro
@@Arnav2322in which class you are bro
@@nkhnouk3781 11th
Ans 3748
You are one of the greatest mathematics teachers as well as more generally,a teacher I have ever seen.........
at 7:30 on screen simplification nahi kiya kyuki hua hi nahi hoga, sir ne 4(k+1)^2 likha tha, vo actually (4k+4)^2 tha = 16(k+1)^2 frame change me 2 ans bhi mark kar diye
Main bhi wahi soch raha tha that 2nd term is not 8 square for that variable😂
mai khud
Sir cengage and other books important questions series le aao for jee 2024 ❤
Khud solve karlo.
@@11111111111111117391 U can take 1 day in a week to do it. Just like a revision day.
I also did same approach but sir failed to write limit.😢
Me who had not seen sequence and series one shot by bhai: *Interesting 💀*
Same 💀🗿
Sir second question can be simply solved by forming a common elements ap upto 10086 and final answer will be 3748
yup
Scrub thypus is caused by Orenntiae tsu tsu gambhusi.... Transmitted by tormbiculid mite.... I'm a medico... Who just knows (a+b)√😅
2018+2018-288
Very simple question 😂😂
6:10 sir yaha par [4(k+1)]^2 aayega shayad
u r right shera.. by mistake i wrote 4 outside the sq...keep it up...best wishes
Wow sir acknowledged it 😊
@@JEEnexus love you sir ❣️
@@JEEnexuslove you sir ♥️❤
Ha mujhe bhi yahi confusion ho raha tha ab samajh a gaya
first qs to nhi aya bhai lekin second qs to 4 min mein correct kar liya idk what is so hard about this qs its pretty basic love you bhai-your shera
I tried half hour then by common term lekar aur phir XUY lekar finally 3748 answers aa gaya
3748??
Sir your teaching technique is very good and easy to understand. I am in 11 and solved the last question.
a=16 d=5×7=35 for x tn=10086 and for y tn=14128 . So the ap formed with first term 16 and common difference 35 will consist of last term less than 288.71. So terms of XnY is 288. Then terms of XuY is 2018+2018- 288 as you directed. The ans comes out is 3748.
How much syllabus is completed bro.. ( I'm also In 11th 🤙)
@@aadityamishra4633 sets, Binomial theorem, complex no, sequence n series, limits, 3d geometry, differentiation, trigonometry, quadratic eq etc. in offline. But at jee level , I completed 6-7 chapters.
@hiphopstan696 which coaching are you in?
@@utkarshsingh9682 I am currently enroll in pw online. In offline I follow practice books like hc verma, cengage and pyqs. I have tution teachers too.
Sorry but I have a doubt
You put 4²+8²+......+(4n)² as
4(k+1)² [0,(n-1)]
But when I substitute the value of k in the equation like when k=0 => 4(k+1)²=4 but shouldn't it be 4²
Similarly, when k=1 => 4(k+1)²=16=4² but shouldn't it be 8² ? 🤔🤔🤔
Well I'm confused if anyone know what I am saying wrong or if there is a mistake in sir answer please explain. Thanks 😊
I think there should be (4k+4)² or 16(k+1)² instead of 4(k+1)². Well I could be wrong in understanding the answer too
ooh sir galti se likh diya oh (4k+4)^2 hoga
@@chetananurag3459 youre right
Bhai 1st time me too nahi bana
But resume karne ke baad aapka pehela step dekha
Uske baad too apne aap kar liya
🥲🥲🥲
But ye samaj nahi aa raha
Ye 2nd wala quesⁿ itne bacchoan ka galat kese ho sakta hai
Esse easy easy quesⁿs too
Mere DPP level - 1 ki starting me hee hote hai
😅😅😅
JEE aspirant
2025
🔥🔥🔥
The question is not actually tough but the main step was a bit intuitive. If you get that step then your question is done. And I actually got this question in my first try. Feeling confident. By the way thank you bhaiyya for the solution.
Sir maine puri series ko 2 halves me toda aur dono ka sum krke Sn nikala to vo product of two consecutive no. aya 2n(2n+1) fr hit and trial lagaya thoda dimaag laga ke and answer comes out to be 37*36=1332 aya pr pahle pr nhi dhyan diya🤧
Bhai soln ka picture bhej skte ho ,, g-drive se link bnake p;lsssss
@@hks8614 wait
@@hks8614 bhai tg id de uspe bhejta hu ya Gmail!!
The best sir:Arvind kalia sir🔥🔥❤. Love u sir❤
This is a direct question from black book :)
Sir I did 2nd question in 2 minutes, thankyou bhai ❤️
Homework ke question ka answer 3748 hoga.And n(A intersection B)=288 hoga .2018+2018-288=3748
Bhaiya ek doubt tha, isme number of common terms mtlb X intersection Y decimal me kyo aa rha hai wo to natural number aana chahiye na??
n
Are bhai second question mein bahut logon ne [x U y] ka answer nahin balki X intersection Y ka answer de diya hoga Joki 288 a raha tha.
3748 ans for the last qn, aaj karke khudse kitna proud mehsoos kar raha hn
Ye toh bahut aasan hai.. Allen module ka q hai.. Kar liya tha 1st attempt mai
question was so easy idk why many people cannot solve it answer is 3748
Sir thankyou for all you have done for us
Bhai, the last question is really easy one. But a tricky one. I took 4 minutes to solve it.
agr is question me AP nhi bnti dikhe
Tb bhi sigma ko negative sign pr break kr k
Dono side ki general terms likh k subtract kr skte isse bhi sum easily aa jati h or phir bs quadratic me value rkhni h
2nd question
Pura sahi se kiya,, last mei galti DIVISION mei Mila🙂🙂🙂🙂
Yes sir the right most term will be {4(k+1)}²
i did it in a completely wrong way but i got the answers in 30 secs💀💀
like i Hadnt considered that negative i oonly saw k^2 and you know that sum of it will be divisible by 6 and when i checked the answers i found A ,D correct but i thought it is wrong bcoz i hadnt considered negative complex term but i am hella completly right
Hello in the first question cant we do by grouping? Like taking 1st and third digit together and applying a^2-b^2 so we get all 2 as common and then its just normal sumation of 1 2 3 4 5...sir please correct me if I am wrong...although I am not getting desired answer from this but my method seems legit
I solved with this approach only you will get the ans
Well you have applied the correct approach just some minor mistakes you have made i think
3748 very easy ..patience needed
Sir i am in clss 9 and i did the second question...i did just by forming the common sequence where a=16,d=35..so a+(n-1)d
Same but I am in 11th
I too did I'm in 11th
We didn't had sets in 9th so i learned it in allen for the first time
are bhai mera bhi 288 par yeh galat tha
kyukin x union y pucha hai = 2018+2018-288
x intersection y = 288 nahi
silly mistake se hi sabka galat hua HAI !!!!!!
@@A-hb5wx I knew and answer is something near 4k
I solved it while solving pyqs but still mere mock me marks nhi aa rhe 😢
Sirrr... Bring black book session and plzz start with algebra part .
This type of questions i slove in dpps🗿 but bhai is best ❤️
Sir my answer is coming 3748.I did it in less than 45 secs (using calculator tho 😅)
it can also be easily done if u pairr like 2and 3 4and 5 a series will be generated
Maja aa gya sir
Sir first wala ques to bhot halwa tha agr aap thoda sa or ache se observe kar lete to
😂😂
Dekho ..
(-1²-2²+3²+4²)+(-5²-6²+7²+8²)....
Matching 1st term with third and and 2nd with 4th and so on ....
applying a²-b²=(a-b)(a+b)...
=>(3-1)(3+1)+(4-2)(4+2)+(7-5)(7+5)+(8-6)(8+6)......(4n)²..
Taking 2 commin from each term
=2(3+1+4+2+7+5+8+6.....)in this sum series every natural number came till 4n...
Applying Sigma 2sigma(sum of natural numbers till 4n)=2(4n)(4n+1)/2=...
Pitting n=9,8,10 will give us required answer..
I think my aprroqch is far better than you sir but I don't claim it my teacher solved it last year in class with us ..I gave due respect tou you sir but I was just giving a better approach for this ques for your students and you too
Love you sir.
Yes this is easier , thank you
please see it carefully bro
i think you have to read the chapter again
@@SudiptaDhar-q5n
Never doubt on Msm sir( Baba)from pw🤐
Now sankalp Bharat...
Best teacher I have seen in my life for maths ..jai baba ki ..jbk jbk..
Agr aapko kuch lga ho ki galat h to pls inform Kari agr hoga to I will ensure it must be corrected else you should see where you find the mistake ,may be your concept is losse or there might be some other reasons that you said that my solution bus wrong...
Galat lg rha hei bhai
Thank you sir❤❤❤❤
Sir dono hi easy Q the
Second ka answer 3748 h
Common ap ke no. Of terms minus krdo total no. Of terms me se simple
Sir last question I got 3749
Is it correct????
sir it is 16(k+1)^2
Physics ke one shots bhi lao sir....
Sir answer 288 (edit-(sorry 2*2018-288) maine intersection le liya tha!!)hoga by observing the equation 5n1-7n2=6
Note- I got this eq by equating the n1th term of AP1 and n2th term of AP2!!
SO/-Isme 5n1 keval 0 aur 5 Unit digits vale no lega isliye 7 ke multiple me ham keval 2 and 7 vale no dekhenge (n2) and by putting values for n2 we get an AP series with First no = 4 and common diff =7 to usse maine 2018 ki position pata ki to vo 288.something aayi to syd yhi answer hona chahiye tq sir🥱
SIR PLEASE PIN THE COMMET IF THE MY OBSERVATION IS RIGHT!! 👍
Nhi bro Q ache se pado
Total terms me se minus krdo 388 answer aa jayega
I got the 2nd question crct bhai❤❤
air 4 2018 got this question wrong, our sir said.
@@GorakhnathSamal oh...😊
288
i wrote 288 as answer but now i realise that i had to subtract it from total terms
Nyce observation
In the first question, i identified the pattern but still wasn't able to do it, as i didn't wrote it the way u did and hence wasn't able to notice that there will be 4 sequences
And in the 2nd one, i got the answer, it was pretty simple ngl, and I'm wondering why so many people weren't able to do it
(-1)^odd/even is never defined.....
Sir small correction 4^2 not 4
KHUD SE HOGYA SIR THANK U
Bhai mai class 10th maharashtra stateboard se padha hu aur ab directly maths karne jara hu jee level ki to nahi hori hai boht demotivation hora hain please start 11th maths in detail on this channel as I LOVE THE WAY U TEACH🙏🙏🙏🙏🙏🙏
Same bro!!!!
@@aryannikalje07 Please up vote my comment so that more students who are struggling in maths can lead justice for STATE BOARD students...
Waha 4(k+1)^2 nahi hoga balki (4(k+1))^2 hoga
Sir physics ke teacher kab aayege
I had solved the second question a 2-3 days ago.
Same
Ans of 2018 question is 3748
3748 last question's answer❤
How do you obtain statistics about a particular question, including how many students attempted it incorrectly and how many answered it correctly?
released on jee adv website when the results are announced
Maine 2nd kar liya ❤❤
yesterday it was in my test and yeah i did ittt
Sir what if we just do sigmak² and use its formula
I am from IIT Roorkee but I was able to solve these questions
Dusra wala sawaal toh nihayti ghatiya tha be jee mains 2022 m unhone isse mushkil puch radhe terms common to an ap wale pattern p
Sir...can you tell me...how are you getting the students answers data???...i want to check them too
😂😂I solved it in a very much easier way.. first me jo series banai h.. uska last, second last and third last terms bhi likhlo by observing pattern. fir series 2nd,3rd and 1st ,4th ko a2 - b2 ka formula lagao yahi pattern follow karo pure series me se 1+3 common aa jayega and bracket me ek nayi ap banegi of 5+13+21...n terms. Solve karne pe 4n(1+4n) answer aata h..
Sir iam in class 11th i attempt 1st question which is wrong but when i attempt 2nd question it was correct and answer is
3748
yes it is correct
yupp ,, u get it correct
sir mene dono khudse sahi solve kar liye 🥰
Aise question to sachin sir rojkarwate hai 😂😂
I did that 80% wrong question correct in first time
me after doing advance calculus integrals involving the gamma, beta, diagama function.. : seprate into sums of odd and even, you just need to check the convergence criteria for which the series converge.. lmao
technical talks
this was tough ?????
i did it in 1 min no capppp
jee 2026 i am cominggg
I did it in 30 secs Jee 2026*
sir in first i tried using a2-b2 = (a+b)(a-b) but not getting it right
Sir I got right answer in 4 min 37 second
Solved in first attempt
Literally this was tricky
Thanks ❤❤
Bhai pehle attempt me hogya 2nd wala🎉
Answer 3747
Sir what happened to vedantu
Mera bhi ek sapna hai ki mera bhi comment viral ho
Sir correct answer aagya 😭😭😭😭😭