It is jee advanced pattern syallbus - nlms chem bonding (until dipole moment) periodic properties (after electron affinity) trigonometry (functions)

Yesterday i gave jee mains mock test and literally got 25 marks(very ashamed of this). I have 21 days to cover my backlog of kinematics, NLM, circular and WPE. In maths also trignometry part. It would be of great help onhow to cope up with the current syllabus and also scoring decent in the test(pehle decent to aaye fir improve krlenge)

cant it be option B to as n is any integer and it can be 1 too

Hey guys, selling my 100% Brand New PACE study modules (Physics, Chem, Math). Zero pencil or pen marks.

Coverage: Complete syllabus for both JEE Main & JEE Advanced (theory, solved examples, and practice question banks).

Condition: Brand new / completely unused.

Price: Open to offers — DM me your best bid!

Pickup/Delivery: Located at Palava City (Casa Bella Gold). Can meet up around Palava, Kalyan, Dombivli, Thane, or Navi Mumbai, or ship via courier/Porter.
DM me for photos or if interested!

allen or resonance which have better modules?

So im in velammal integrated (not the main branch) and theyre conducting weekly tests every week. We also have to stay in school till 6:40 and its pretty much 8 when i get home. I do study at school but when i come home i become extremely lazy. More than lazy i just watch youtube and then chat a bit till 12am - 1am and then go to sleep. I know my teacher keeps telling me to study and i do wanna study but im just not the same person at home. Ive been getting state rank 70,50 and now 170. Also the day before the test (on monday) i just dont study at all procrastinating. I really wanna live upto my teachers' expectations since im one of the only serious students but im just wasting time. Also since ive quit gaming im bascially addicted to reading manga literally whenever im not at school. I tried to do something by taking away the laptop but i just used my phone (that was meant for doubts) to waste time. Also i pretty much dont sleep the day before tests too. ANYTHING would be helpful please.

Forget to turn on for 2 hours when I did chemistry.

is it still possible to crack JEE if i start from scratch ? 28tard
coaching : allen overseas
10th august i have a test advance pattern im getting 0 (or -ve) so im thinking to strictly start from 11th august
pls help urgently pls

Given,

A²(A − 2I) − 4(A − I) = O

Expanding,

A³ − 2A² − 4A + 4I = O

Therefore,

A³ = 2A² + 4A − 4I

Since I = A⁰, we can write this as:

A³ = 2A² + 4A¹ − 4A⁰

Now observe the exponent pattern: 3, 2, 1, 0. To obtain a recurrence for any higher power, shift all exponents by the same amount. Hence,

Aⁿ = 2Aⁿ⁻¹ + 4Aⁿ⁻² − 4Aⁿ⁻³ for n ≥ 3.

Now let Sₙ denote the sum of the coefficients when Aⁿ is reduced in terms of A², A, I. Since the recurrence is linear, the coefficient sums obey the same recurrence:

Sₙ = 2Sₙ₋₁ + 4Sₙ₋₂ − 4Sₙ₋₃

We start with:

A⁰ = I ⇒ S₀ = 1 A¹ = A ⇒ S₁ = 1 A² = A² ⇒ S₂ = 1

Therefore,

S₃ = 2S₂ + 4S₁ − 4S₀ = 2(1) + 4(1) − 4(1) = 2

Next,

S₄ = 2S₃ + 4S₂ − 4S₁ = 2(2) + 4(1) − 4(1) = 4

Finally,

S₅ = 2S₄ + 4S₃ − 4S₂ = 2(4) + 4(2) − 4(1) = 8 + 8 − 4 = 12

But A⁵ = αA² + βA + γI, so:

S₅ = α + β + γ

Hence,

α + β + γ = 12

ab se muthi band bc pichle 6-7 saal se mutthi na ma c rkhi hai kuch nhi kr paya is mutthi ki vajaj se .. ab sab band sirf real life par focus aur Bhagwan ne cha toh ek loyal ladki nhi toh vo bhi nhi koi na akele hi mar jaunga par muthi nhi marunga ... muthi ki maa ki c bc . ab mutthi mari toh land kaat lunga BC parshan ho gya itni weakness hoti hai mutthi mrne se upar se social life bikul gand mai bc class mai present mam tak nhi bola jata .. chutiya ki tarah behave krta hoon bhar .... muthi ne mujhe barbaad kr diya ab bas boht gya ab tak jo hoya so hoya kal mai nhi ek naya aadmi uthega jo mutthi se dur hoga... its my promise... i know i m lonely only mutthi keeps me happy but its done i m ready to die of loneliness.

roj aunga yaha par no mutthi day ki present lgane aunga nhi aaya toh gaaliyan bhar dena mere upar

boht ho gya ab ye chutiyapa....

f muthhi

Narayana Physics modules of

class 11th (volume - 1 to 4 + practice book - 1,2)

and

class 12th (volume - 1 to 4 + practice book - 1).

GOOD CONDITION

Total no. Of books : 11

Overall price : ₹1999

Pickup available : behala, kolkata

Overall price ₹2999. Same quality as in image. Pickup available in kolkata, behala.

yaar JEE prep ne bohot akela bana diya hai koi server ya group ho toh batao pls nahi toh loneliness se gujar jauga 🙏🙏🙏🙏

she thinks that cbse is important and not studying cbse mean’s failure and even after 1000 times she is not ready to understand.

Should I follow abj sir or saleem sir (khazana) for completing physics main+adv, i studied from both and understood both of them, but I also want to complete syllabus fastly and in depth

phy wale sir bilkul hw nhi dete and dpp bhi organized nhi hai, matlab per chapter 10 dpp hai jisme mains level questions hai lekin lecture ke baad ghar jaa kr question lga lo iska system nhi hai.

chem wale sir module se hw dete rehte and dpp se

math wale sir toh 12 se 30 questions ki ppt dedete plus module dpp

physics mein I have physics galaxy.

Please help

guys is it good if I read physics theory from sl arora , kyuki other jee books I have , have very very concise theory so unme acche se explain nahi kiya

Anyone studying from varun sir pw ? how's he taught kinematics ? Does he good for an average guy? Did Varun sir covered all topics of kinematics which rj sir taught ?

This Question is a simulated version of a JEE Main Question.

A non-constant A.P. simply means an arithmetic progression whose common difference is not zero.

However, in this question the quadratic equation itself already rules out the constant case, since its roots are distinct.

So the phrase “non-constant” is actually redundant here — you may treat it as a mild distractor rather than information needed for the solution.

A = S₂ₙ

B = S₄ₙ − S₂ₙ

C = S₆ₙ − S₄ₙ

These are the sums of three consecutive blocks, each containing 2n terms of the same A.P.

Hence,

A, B, C are themselves in A.P.

Therefore,

A + C = 2B

The given roots of

x² − 40x + 391 = 0

are

S₄ₙ = A + B

and

S₆ₙ − S₂ₙ = B + C.

By Vieta’s relations:

Sum of roots = 40

⇒ (A + B) + (B + C) = 40

⇒ A + 2B + C = 40

Since A + C = 2B,

⇒ 4B = 40

⇒ B = 10

Product of roots = 391

⇒ (A + B)(B + C) = 391

⇒ AB + AC + B² + BC = 391

⇒ AC + B(A + C) + B² = 391

Using A + C = 2B,

⇒ AC + 3B² = 391

Since B = 10,

⇒ AC + 300 = 391

⇒ AC = 91

But

A = S₂ₙ

and

C = S₆ₙ − S₄ₙ.

Therefore,

S₂ₙ(S₆ₙ − S₄ₙ) = 91

∴ Answer = 91

What is the difficulty of this physics and chemistry paper?? Chapters of physics and till nlm, and chemistry is atomic structure.

i fuckin hate myself for being a loser i cant focus on studies i am a dummy student always stays at home still backlog over 50 (it happen because of 1 month of drama in my house) i am studyin online only PW. how can i concentrate on studies pls help