JEE Advanced 2023 | AIR 1 type question, hardest in the paper

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Time Stamps
00:00 Analysing the question
00:50 Solving
09:50 Final Learnings
10:17 Set of 60 promotion

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[@s.j.745]

i solved it in 2 days🙂

[@hacked9032]

As a guy who cracked Neet.. good thing i didnt take maths 💀. 1000 page ki books ratt loonga but yeh mere bas ke bahar hai😂😂

[@padmasangale8194]

*"Bruh why all the people in the comments are telling this was easy and even some are telling that they solved it orally (sarcasm)💀☠️. Bro ur whole comment is filled with toppers having AIR

[@priyanshu_7446]

Un sb ki m k c

[@deepanshu-verma89]

😂😂

[@godson200]

aree ye sab fekte bohot hai. asli paper ka reality yahi h 50% marks bhi score nahi kar pate 99.8% of jee advanced aspirant that's a fact you cannot deny

[@godson200]

par after solving the question, it was actually really mains level question ngl. maybe exam pressure hoga baccho pe

[@ashishgope8211]

@@godson200 aa gye aukat pe , itna pravachan dene ke baad

[@expcollector7215]

I did it this way: first found out a general term for S by observation as: 7*10^r+1 + 5*10^r-1+....+ 5*10+7 for a general 'r'. then just apply summation of this from r=0 to 98 and use gp formula on 10^r+1 and compare with the other form to get m and n. this is the most basic approach that can be thought of during exam i think, that 10S one is too intuitive for me.

[@Realsxullz]

same, thats what I did as well

[@nimmubhai4779]

same

[@Shweta-zs5jr]

instead use general form 7x(10^n) + 5/9{(10^n) - 1} + 2

[@akshatbansal7622]

Same

[@RajeevRanjan-tp1ku]

​@@Realsxullzsame bhai Maine bhi aise hi Kiya tough utna nahi tha isse 6 min me hogaya 10s wala method ko apply karne ke liye 13 ka table acha se yad hona cahiye tab hi wo click hoga ki Haan aise bhi kar sakte hain

[@math_Nerd-nt2gb]

What I understood from the question is that when the terms of the series are increasing digit by digit you should multiply by 10^(no of digit increases per term)

[@a--p]

I solved it with a completely different approch jese bachpan me likhe the wese 77 + 757 +7557........ likha then isme pattern notice kiya ki is likh sakte hai as 7(99) [ones place ke sab 7 ka sum] +7(10+100+1000.....10^99) [Remaining 7s according to its place] +5(10+100+1000.....10^98) [Remaining 7s ke just right wale 5s] +5(10+100+1000.....10^97) [just right wale aur 5s] . . . 5(10) [aakheri tens place wala 5 jo series ka last digit hai] then (10+100+......) wale part me GP ke sum ka formula lagaya 75...99.....57 wala term dikh gya +57 -57 karna para kyuki initially 75............500 form hua tha then answer 1219 aa gya {First attempt me starting me 7(100)+......... [100 nhi 99 hona tha] kar diya tha ans 1282 aaya tha fir video dekha toh pata chala ans 1219 hai thora soochne ke baad galti dikh gai ki 7(99) +7(10+100+.......) hoga then correct answer aa gya }

[@4fgaming925]

hamare sir ne same method se karaya nice bhai

[@user-jx1yf7yg5d]

Chutiya tarika hai yah 😂 bas calculation karta jao jao orh time waste hai orh kuch nhi . Rather than Asa karo S=77+757+7557+. N Multiple by 10 on both the side 10S=770+7570+75570+755570. 755..............570 ​Now subtract both then you will observe a pattern 9S=-77+13+13+13+13+13+13. N [phone].........70 Now 9S=13*98-77+7555555.......570 S=13*98-77+755555555.....70/9 Hence M+n is 1219@@4fgaming925

[@satyasrilakshmi6626]

I split the term and got some value, pretty sure its correct but getting it in terms of the question was hard, nice method bro

[@user-rt4nu1fd4f]

Bhaiya I swear diwali ki chutti me yehi sawaal kiya tha 1st try me ho gya muskil nhi laga tha ☠️☠️

[@Ayush-mg6xw]

Bhai ye Pathfinder ka sawal hai mai karta tha fun mein bachpan mein uska solved exam hai usme method tha 10 se multiply krdo aur subtracr krdo fir gp ban jayega 2 min mein ho gya mujhe fix approach pata tha aisa mere saath kash mere paper mein bhi ho jaye that's why reading solved example are a must

[@piyushjain7687]

Bhai mei aaise Salwa 6th class se karta aaraha hu

[@Ayush-mg6xw]

@@piyushjain7687 average jain 🛐🛐🛐 bhai jain aur agarwal kis mitti ke hote hai I can trust anyone easily mai toh tomar hu naam bhi nahi ata kisi ka top 1k mein bas achi baat ye hai I'm obc

[@piyushjain7687]

@@Ayush-mg6xw majak kar Raha hu Mari preparation mei bss physics ke L Lage hai 😣 pathfinder try karne ke liye base hi nhi bann pa Raha hu maths badiya hai 🥵🗿bss iss questions ko try Kiya tha first attempt meaning bana tha because GP padhne ke direct baad iss question pe kud gaya tha

[@Ayush-mg6xw]

@@piyushjain7687 bhai tu jain hai under 500 pakka

[@ayushmishra1809]

thank you bhaiya...doing every lecture

[@jeesimplified-subject]

good going bro.

[@smitpatel5107]

Bhaiyaa, kattai jeher sawal hai... Karke maza aa gaya... 😎😎

[@abhinav-ed1ui]

the gen term is nth term of GP 7.10^(r+1) + summation till nth term of GP 5.10^r + 7 Then we just have to sum it all up, again there are some GPs and relate it with the given eqn.

[@cubingdom3426]

Bhaiya ek time tha jab aapke videos dekh ke dar lagta tha, ab to comments padhke aur bhi jyada dar lagne laga hai

[@shaikabdulrasheed4414]

😂😂

[@AdhDumy]

Bhaiya , I need to ask one thing that even if answers humare aa rhe hain , it takes a lot of time... like 14 - 15 mins lg rhe solve krne mein... but advanced ke paper mein itna time lga denge fir toh mushkil ho jaayegi.. time required to solve the question (**that too with alertness**) kese kam kre?

[@swanandpande9338]

I first found out the general term of 75...57 series and then calculated the value of S using finite sum of GP. The only time consuming part was to relate S with (75..57 +m)/n

[@user-uh7vo9ff4z]

Bhaiya question bahu accha laga aur jis tarah apna explain kiya question easy ho gaya

[@Notsosarcastic_02]

It was quite easy , the method was simple , the catch is multiplying S by suitable number to get the answer

[@user-ts1uw6rm4j]

Bro but us time par ye sochna bhi to hard hai , ye waise board level par bhi ata hai question

[@Notsosarcastic_02]

@@user-ts1uw6rm4j yeah you got that right . But by practicing more and more and staying calm really helps you think out of the box

[@user-ts1uw6rm4j]

@@Notsosarcastic_02 so you are in 11th right

[@Notsosarcastic_02]

@@user-ts1uw6rm4j i am a dropper 😂 scored 96 in jan

[@indusarrow9605]

I m not able to solve all the adv questions myself but did it really boosted my confidence great question I justed multiped with 10 and shifted the sum than breaked the last term into given

[@dhruvm.s.1103]

Bro diffn in GP se nahi horaaa ... agar kisi ko ho gya plz batao

[@psychobuddy4122]

the first stuff that clicked in my mind was ki pattern based hi sum hai...then deemag me aaya ki add subtract multiply krke kuch to hoga s ke sath n un dono ko minus krdenge jisse ki hamara denominator ayega n wala but upar wala smjh nai aa rha tha...pehele 1st 2nd and 2nd 3rd ko multiply kia dekhne ke lie kuch aage wale terms to nai ban rhe...nai bana...fir socha 10 se multiply krdu? then minus krne gye to -77 and 13 dekhkr laga nai..repeat nai ho rha hai ..age nai kia and tha ki last me ajayega so usko kaise hataungi...badme minus wale isse gp ka aaya..usse ban gya ...lekin was not sure..to numbers ko divide krke check kia 9.8 pehele ka then 9.83 smth aaya tb lga answer dekhna chahie...bht khushi mili ye dekhkr ki dono methods sai the😭 edit: sb no ko expand krne pr kuch patterns banenge? for eg 75557 ko 70000+5000+500+50+7??

[@ashutosh4189]

Bhaiya i tried this que mera ans last tak shi aya but ek choti si silly mistake ( 1 ki jagah 2 ) rakh diya flow flow mai😢😢

[@drdripransom342]

I was the only one in my coaching to have solved question like this alongside my teacher (obviously), in questions like these where some kind of recurrence pattern is forming, this AGP method works usually, a few terms will first trouble you but with enough practice it becomes easy So trust yourself and work hard

[@asutoshpujapanda8665]

great solution🙌

[@Freak_GTA]

10 multiple wala kaise hit hoga agar pehle nahi banaya hai to

[@krsna86537]

Cengage ke illustration me tha same ye question

[@vapix7561]

Maza aagya brilliant solution

[@rudrakshayadav277]

After aadha page ki ghich-pitch I was able to think of this method and did it correct. I always thought my maths was weak but solving this really boosted my confidence. Now preparing for advance for 26th

[@priyanshichauhan3409]

THANKS BHAIYA

[@amitotc]

I have not seen the solution (writing this comment after pausing the video). It is very easy to see that there are numbers with digit repeating which can be compactly represented as GP. There is yet another sum running over different GPs with different first term and last term. Kind of double summation which needs some manipulation to arrive at a compact representation using another GP sum which when reversed will give the repeating pattern number along with some constants m and n. I don't know if this is the solution described in the video (but I am taking a guess by just pausing and thinking about possible solution). It took me less than 30s to think all this. Does not seem like a hard problem at all.

[@kushaankoul5775]

hi can you tell me how you get these stitistics of how many people attempted and and correct

[@dhanushnayak]

search for jee advance report, you will a huge pdf, go to question paper analysis in that

[@tilakagarwal8163]

I was able to solve it in under 10 minutes using a much simpler and obvious approach. You can try adding up the unit, tens up till ten thousandth digit by adding 7*99 + (5*98 + 7)*10 + (5*97 + 7)*100 and you get 61683 as the last five digits. You know the sum must be of at least 100 digits so remainder “n” can not be greater than 12 otherwise the numerator would be greater than 101 digits which shouldn’t be possible since m and n are less than 3000. Then you can multiply the last 4 or 5 digits with each number up to 12 and will easily be able to note when you get to 9 that it’s the only time the multiple is greater than the sum in the threshold of 3000. You can continue to check a few more digits when multiplying by 9 but will see it easily and see the multiple ending in 6767 which is 1210 more than 5557 and you get 1219.

[@dakshsingh5891]

Too lengthy

[@tilakagarwal8163]

@@dakshsingh5891 True, it do be quite lengthy but simple enough to think in a minute and can still be done in 5 minutes

[@formaniclv1046]

​@@tilakagarwal8163yeah bro.. I solved it in the same exact way itself..it's not that lengthy though

[@user-jx1yf7yg5d]

​@@tilakagarwal8163still lengthy. Rather than multiple S by 10 then subtract Then S will become 9S=13*98-77+755555....570

[@tilakagarwal8163]

@@user-jx1yf7yg5dbruh at that point you’re just guessing the remainder to be 9 and hoping to be correct, it’s like saying why don’t you hit in the first trial of hit and trial

[@priyansh8962]

Bhaiiya easy laga tha....practice itni karli hai sequence ki...Sum dekh kr hi idea laga tha ki T2-T1 wala method lagega

[@Dalal_The_Pimp]

Sameer bansal definite integral Ex 5 ke Q.50 me ek result tha ki (111...11) n times= 10^(n) -1/9 waha se sochke pattern bana liya (755...57)=(777..77) (n+2) times - 10 x (222..22) n times, iske baad bas calculation aur consolidate karna tha.

[@annapurnakushwaha2859]

Yup

[@jeesimplified-subject]

aree bhai bhai 🙌🏻

[@a.sthakur3433]

same aisa swal kr chuka hum prove that n(4s)n-1(8s)9 always form q perfect square

[@_Longwinded]

Left JEE preparation 6 years ago and got this video in feed, managed to solve it after some time but I solved it from different method than you suggested. Yeah I felt proud solving that question. Also I can share the method. But I don’t think I can type it here.

[@rawatutkarsh]

Another way of solving this question is by using method of difference which a lot of people don't really know about. If you notice, you'll find out that the common difference between consecutive terms is in GP. Therefore by using method of difference the general term can be given by a(r)^n-1 + b where 'a' and 'b' are variables, r is the common ratio of gp. Using n=1 ans n=2 you can find the value of a and b and thus you'll get the general term. Now just use sigma on general term and do some basic manipulation to get the answer. This is how I solved it during my first attempt. Edit - didn't see bhaiya already mentioned it in the video. I just commented without seeing the video

[@SANDRA-xq7es]

To delete maar abhi.. :)

[@rawatutkarsh]

@@SANDRA-xq7es 😢

[@vyukastra]

​@@rawatutkarshBhai fir bhi sirf 424 bacche solve kiye . I mean ye out of the box ques bhi nhi tha only bhaiya ka method was like too out of the box . Actually me exam situation kuch level of difficulty badha deta hai or else Ghar me baithke toh kitne saari ne solve kar diya

[@rawatutkarsh]

@@vyukastra sahi bol raha hai bhai Coaching tests me bhi yahi hota hai simple sawal time pressure me mushkil lagne lagte hai

[@creamfied1681]

thank you bhaiya at the beggining i paused and tried it, it is an easy question unless you dont know about agp

[@munimahmed8489]

I had this question solved in jee advanced 23 but still I couldn't even qualify my maths cutoff got 6/8 🤡 baaki subjects ka hogya tha. My sequence and series was good I guess.

[@Bumseidon]

Maine t2 aur t3 number tak sum likh ke us format mei likha aur fir mujhe pattern dikh gaya , m was in ap with 13 ki difference and N constant rahega wid 9 😊😊

[@criticalthinker9134]

Just takes around 5 minutes to figure out n=9 using either - Principle of Recursion - Asymptotic analysis - Linear algebra ( eigen analysis)

[@tanupatel7588]

konsi class me ho ?

[@utkarsh3945]

chod liya gyan ban gya hero?

[@braek2766]

@@utkarsh3945 It's a basic question testing intuition, Taught very early in advanced sequences and series.

[@Ghostwriter_zone]

& for m We need to solve collatz conjecture😂😂

[@user-lv1jk9qb9t]

How did u do it by linear algebra.. Can u explqin more

[@Deadlox2012]

Bhaiya ji Aasish sir ne padhaye to tha but Maine dhiyan nahi diya.

[@pranavkapsimey6393]

The final S had 99 times 5 but you only multiplied by 10 then how it can be 75......57(99times 5) it only had75......57(98times 5). But if something I'm getting wrong please correct me with that.

[@parjanyak1104]

After multiplying with 10 the last digits became 570, then he split it as 557+13 , where we get another 5. And got a number with 99 5's. Check at 9:26

[@JohnDoe-hl2qm]

​@@parjanyak1104👍

[@pranavkapsimey6393]

@@parjanyak1104 got it.

[@abhi.5610]

Maine to almost last step tak kar liya tha aapke method se last year 2023 jee advanced me...but last step me mistake ho gaya tha or answer galat aaya tha

[@karajkot6735]

I did break 77 = 22+55 757 = 202 + 555 7557 = 2002 + 5555 Now 2002 = 2(10^3 + 1) 5555 = (5/9)(10^4-1) And that's how I solved it, but it was still calculation expensive.

[@NidhishwarReddy]

sir mene i solved this today morning what i did was write 77 as 70+7 , 757 as 700+50+7 , 7557 as 7000 +500+50+7 .... which we will become s = 7*99 + 7*(10+100+1000....) + 50( 1+ 11 +111...) phir isko solve kara aur ansar agaya roughly 8min lage the

[@Shreyanshnamdev07]

Same mene bhi Yahi socha tha pr mene aage ki calculation nhi ki kyuki mujhe maloom tha ki ye general approach h to answer aa hi jayega BTW well done !!❤

[@KushalChaturvedi-ng1iv]

I have solved and my answer is 1292, let's check what's the answer Edit:I checked the solution and i have done silly mistake in calculation ans is 1219

[@truth1549]

wow topper lmao

[@Truthprevails852]

i did it by the longest and most calculative method possible , took me 25 minutes , i just kept on solving and was about to leave it

[@user-jc6sk4bo4v]

did it with a different method by splitting the terms lots of calculation error par ho gaya

[@d.m.b.2836]

ek approach ye aayi thi - sabhi do consecutive terms ka diff nikalo usme se 68 common lelo gp aaegi uska sum easily aajaega. thus now we have S and uske baad easy hi h....

[@deepanshugupta1987]

Question Easy Tha But Mere Solution, Ne Hi Mujhe Uljha Diya

[@Rishabh-wc6fv]

One other method is that - notice that difference of succesive term is in gp (First term 680; r=10) we can simply apply Tñ=arñ+b

[@Rishabh-wc6fv]

I think most simplest method

[@piyushsharma18941]

I did with that but it give the result tn-t1 = sum of that gp How to proceed further

[@rawatutkarsh]

That's how I did it too Method of difference when common difference is in gp 🙇

[@hrishinaik281]

Bhaiya where do you find such statistics about the question ?

[@jeesimplified-subject]

Official website of JEE ADVANCED

[@kushgaddamwar1480]

merse solve hogya sir. bhot maja aaya solve krke

[@utkarshranjan5705]

Bhaiya what a great solution

[@sahilagarwal8335]

In hindesight its good to take this approach, but to start visualizing , begin by expanding number , automatically the AGP form will click in brain. Its a simple one though. Things with double summations over two variable would eat up the mind more

[@ruchikaa_ray]

I did this but with a complete different approach like by replacing each term of S with 99 repetition wali terms

[@banerjeeproductions4625]

Bhaiya cengage ka ek sawal isse match karta hai... Exact nhi hai lekin us question ka concept use hua...

[@anmol4205]

Question asaan hai but needs a little bit of patience and calmness !!! ❤

[@srajansingh-ql9kx]

Bhai yeh sawal literally jab solve kiye tha bahut aasan laga tha kyuki 1st try me aise hi kiye tha abb jab video dekh rha hu pat chal rha uss paper ka hardest sawal tha

[@Arnav1883]

I am in class 10 and somehow found the approach in around 1 minute. I used S=77+770-13*11+7700-13*111 and so on. writing s like this i was able to find the value in 4-5 steps and then using same observation on 755...99times7 i got the value of m and n :). I am still surprised i was able to solve it cuz I usually get stuck on jee adv. pyqs

[@C__YashKumar]

I have solved this in 3rd try due to calculation mistake. In my method, I have expanded the number in powers of 10, uske baad to sawaal kuch bacha hi nahi.

[@ParthBnsl-iitis]

I was able to solve the question but most probably becoz I have solved it once previously... I used 10S - S wala method, and also figured out that no. Of digits badane hai toh 10 se mult krna hi hoga, I think that's the only clue/hint that is required... Also I brainstormed almost an hour to figure out some other method but wasn't able to figure it out... Bas itna nikal paya ki m+n=some odd no. Hoga... Also I tried ki m S ke last digits nikal kr m aur n pr comment kr pau but it was a very poor way... Also I was expecting this same soln, nice one👍

[@kunalmehta3791]

Can be done by writing expansion don't take that 99th term, write all the other ones in expanded forms can clearly see that the 5's and 7's Term of expanded form form an agp solve them but this becomes a bit calculative

[@jeesimplified-subject]

ohh, I see 👀

[@user-jx1yf7yg5d]

This shows lack of practice dude it is probably the worst method to solve such kinds of series

[@mayankchawala6209]

Cengage me bhi ek illustrartion hai iska

[@Abhiviews]

Math is beautiful!

[@boomer1248]

Very fun, i did it with 2nd way, after taking 10 mins

[@aayushmanchaturvedi3706]

I remember solving this question for IOQM last year in class 9th This was one of the only question i could do, and for me this was one of the easiest The problem is so many people get so absorbed in the advanced topics that they forget the basics of it all

[@emmawalker5461]

So true

[@PradeepDubey889]

Bhaiya mujhe ye question 4th try me ho gya I am im 11th. 😊😊

[@aryansinha4992]

bhai ye stats kaha se dekhte hai?

[@user-tc3gq9pl6p]

jee adv report

[@siddhichougule]

Aj dekhne Mai thoda late ho gya boards ka kr rhi thii🥱😂btw I solved the question half✨🔥

[@BEEAHMADNASIM]

Use recurrence relation ... More smart way of solution....

[@avishmakar8641]

ALT APPROACH CLASS 5 TH METHOOD _____ WE CAN WRITE 77 = 7 .10@0 + 7 .10@1 here @ means power............ similarly we can write T(r) = 7 into 10 to power 0 + 5 into 10 @ 1+ 5 into 10@2 + 5 into 10@3 and so on.... + 7 into 10 @(r+1) . here this will GP and we can use sum of GP AND REWRITE GENRAL TERMS T(r)= some expression in ( r) and we can use sigma and it will get solved by doing some manipulation

[@sss-nd3ol]

can u pls explain this method again? thank you so much!

[@avishmakar8641]

@@sss-nd3ol terko yeh to pta hoga ki hm kisi bhi num for eg 77 ko 7 into 10 ki power zero + 7 into 10 ki power 1 is trh se rep kr skte hai SIMILARLY mai yha power ko * SE REP KRUNGA TO General term T(r) hai 7555........7 isko hm 7.10*0 + 5 .10*1 + 5.10*2 and so on 5 .10*r + 7.10 *(r+1) likh skte hai ab dekh yha 5 wali term main gp hai to formula use kr le or genral term aajaygi T(r) usme ab sigma aplly krnege to sum miljayega or hme que mein S= T99 +M /N NIKALNA AGAR DHYANN SE DKEHE TO TODI MANIPULATION KRNI PDEGI BUT HOJAYEGA

[@abhirupkundu2778]

hmm, not bad.

[@howdadogdoin729]

I did same :D

[@ink.indkan]

I remember doing this problem back in IOQM prep but X boards are holding me down.

[@user-wu8il8qb7r]

amazing

[@nil_1234]

I had a similar thought like the second approach but, I mixed it with the first one 😅 that's why my answer was not coming😅

[@yatharthupmanyu2228]

Mtlb apan terms ko Tn= 75......57 =75........ 5+2 , kuch aise likh lete hai aur uske baad Tn= 7×10^(n+1) +5[10°+10¹+10²+10³...... +10^(n) ] + 2 ,, bs S= sigma Tn normal se khud se ajaye aise hi manipulation hain T99= n S- m, mai leke ana hai,, and done

[@Immartian18]

Bhaiya maine toh general twrm nikal li 7. 10^r +5/9 .(10^(r-1) -10) +7 Or puree pe sigma laga diya gp aue ap mai tooot gaya question

[@satendrakumar7281]

Sachin sir ne ye hi method bataya tha esi series ko solve karne ke liye is that's why solution dekhne ke bad kuch khas nhi laga but me ise exam me paka chod deta because mere se series hi nhi banti

[@nishantsaini_13]

Bhai maine to aise kia ki each preeceeding term mai 10 ka multiply karke 13 minus karde to next term mil jayegi. Isse maine general term nikali 77[10^(n-1)]-13{[10^(n-1)-1]}/9 Fir maine isse solve kiya aur ye answer aa gaya

[@thesecretartist327]

Apne answer ko please thoda elaborate karna ki aapki general term kaise bani?

[@yashmaheshwari9908]

Yeh method se to mene bohot easily Solve karliya.

[@Dr1z.]

I feel i have seen a similar question somewhere, literally only the values were different. And it was before 2023 adv paper... So probably the book already had a qs like this??

[@arindam3276]

Bhai I think tum uss wale ki baat kr rge ho jo cengage me hai S=7+77+777+7777+....

[@Dr1z.]

@@arindam3276 Nope, aisa ek qs mere notes me hai...

[@SleepyAurora-wq6xl]

Sir how would we know that we have to minus 10S? How can I develop this kind of intuition?

[@godson200]

bhai terms ke beech ka difference is in GP. difference between terms are 68,680,6800. clearly the ratio is 10, ye toh commerce ka student bhi bata dega. tum log jee ke liye padhte bhi ho?

[@SleepyAurora-wq6xl]

@@godson200 thanks dost

[@SwastikMishra-kn2bv]

I did it bhaiyya in 20 min by splitting terms 2 and 3 times I focused the most in maths during 11th (now going to 12th)at each and every topic that is helping me now to solve such problems

[@djshadowpo2.023]

Mam koi tips for juniors entering into 11th?

[@BraddysReagent]

@@djshadowpo2.023 be consistent either study 3hrs or 8hrs but study daily and revise everything u study on that day only

[@WantsPeace]

Bhai 10th me hu under 10mins me solve hogya aur bilkul same solution tha mera or apka ❤

[@WantsPeace]

Hlo

[@a.m.videos3254]

Ok... tum hi laoge AIR 1 JEE 2027 Mai.

[@blaareen.]

@@a.m.videos3254 bro 2026 mein

[@a.m.videos3254]

@blaareen. Maine assume Kiya ki uska session just start Hua

[@WantsPeace]

11th me hojaonha iss saal❤​@@a.m.videos3254

[@AbhishekKumar-pg7zl]

Iss level ke sawal sachin sir class mein karwate the issi method se par main mand buddhi wahan bhi aur yahan bhi apni buddhi lagane par pada hoon..aine aise kiya ki s=sigma 2 se 99 (Tr). Isme (Tr) is sigma 7(10^r-1) + sigma 5(10^r-2) + ........7sigma(10)😅

[@shaziausmani9598]

By watching this series, i have learnt that almost none of the adv questions are difficult when solved with calm mind at home Solved this one also but this time it got a bit lengthy and spent approx 10-15 min💀 for the answer to come in the first attempt

[@prismatic-bl8qf]

"solved with calm mind at home" yeah they dont give you that at the exam centre. Jee is hard for a reason

[@shaziausmani9598]

@@prismatic-bl8qf wahi baat mein point out karna chah raha tha Examination temperament and confidence on preparation are very important

[@curiocitizen1069]

I solved it in 3 minutes! I am from IIT Roorkee, now living in US

[@curiocitizen1069]

@@666Lindane I am pursuing PhD in Aerospace engineering

[@rishivalluri714]

I solved it because I was already taught to attempt this type of questions.

[@lohithg1458]

Bro me and my friend solved this question and it took us almost one and a half hour to solve it 😊

[@antagonisticalex401]

Aah, the classic, the timeless, 10Sum minus 1Sum trick. Truly one of the staples and favourites of Sequence and Series questions. It is one of the most common, important, and crucial methods used in higher level Series questions. Thats why doing it was my first intuition as well.

[@jeesimplified-subject]

u a jee aspirant?

[@ayushpandey8963]

Same here my intuition was same as well but i didn't solve it lol!

[@ruparelvishesh7194]

I solved it using general term method and it took me around an hour as i did so many calculation mistakes because of how underconfident i was that i can't solve a hard jee adv question, but i am glad i did solved it correctly at the end! At last i felt like it was just a 5-10 min question if done without errors, but koi na i will work on my calculation! Any tips on how to improve it?

[@annapurnasarangi9999]

What was the general term that u got?, was it like summation of (r+2) from 0 to 99?

[@Ayush-yj5qv]

Agar 11th mein ho to har ek sawal time bound karo nahi toh mere jaise L lag jayenge jee mein mai class mein topper tha maths mein bohot acha tha lekin jee ke time revision ke chakkar mein practice chut gya time bound uski vajah se ye haal hai

[@Shreyanshnamdev07]

I think ye 5th class ki expanded form wali method se bhi ho jata let's see S= 70+7 + 700+50+7 + 7000+500+50+7 +______+ 7×10^99 + 5×10^98+(___)+7 then next step we simplify 7^99 + 5(1+11+111+1111+_____+11.....111) + 7(10+100+1000+____+100000.....) uske baad wahi 11th class ki general approach thoda calculation hogi.... but ansewer aa jayega BTW apka method jyada easy h 👍

[@mohitbang3117]

bro time limite hota hai

[@yesiplaygames5986]

Sir kasam se pehli baar try Kiya jee advance ka ques. Pura 10min mai nikal gya. Khud vishwas nhi ho rha bc kaise nikal gya😅

[@yesiplaygames5986]

First add unit place 7 i.e 7×99. Then find 10s. Now 10s-s and the answer will be visible

[@user-qt9sl7st5h]

Mene pura alag alag krke 3 series bana li ...calculation zyada thi but ho gya

[@Premium_Youtube_User]

As a JEE ADVANCED aspirant normal question jaisa h ye question h ye waise hume kal hi class me krwaya gya hai

[@gidskdsfjiafjifdifjdif]

Bhaiya at first look i multiplied the S with 77 and that was my mistake

[@HangingQueen]

Actually not too tough question. But exam pressure me karna mushkil h for sure

[@jatinmohabe5986]

Bhai mains level ka question ka answer check kar raha tha ya dakh liya pura par ma shayad ready nahi hu advance ka liya 😅😅

[@phoen1x900]

Bhaiya ho toh Gaya Answer bhi mil gaya magar yeh sure ki yeh time killer question hai 10 mins se zyada lag gaya but maza aaya karne mei Actually hum log na eg 755557 ko 777777-10(2222) aise likh sakte hai Aur last step mei m aur n ka value lane ke liye basic manipulation karne laga

[@neetudhakar2797]

Solved in 2.5 mins (inclusive of reading time)

[@phoen1x900]

Air 0 ho ka?

[@sniperkartikyt7748]

TBH i solved it after brainstorming of 20-25 minutes but this soln was quiet good😊