🚀 Sequence & Series JEE Advanced PYQs🔥#jeemaths #jeeadvanced

🚀 Boost your JEE Advanced preparation with key Sequence & Series questions. Watch this short video to understand the most important PYQs, broken down for you in an easy way! 💡
#jeemains #jeeshorts #maths #iit #iitjee #jee

[swetapadmamahapatra]

Option C is correct. Answer is 0<x<10.
Given: the sum of infinite G.P = 5.
First term = x
a/(1-r) =S (for an infinite G.P), where a is the first term and r is the common ratio.
Substituting x and 5 in the equation, we get,
x/(1-r) = 5
=>x/5= 1-r
=>r= 1-x/5.
For an infinite G.P, |r| <1
=> -1< r <1
Substituting r with (1-x/5)
We get, -1< 1- x/5 <1
Adding (-1) to each term we get
-1+(-1) < 1 - x/5 + (-1) < 1-1
=> -2 < -x/5 < 0 [ multiplying each term with 5]
=> -10 < -x < 0
=> 0 < x <10.
Thank you very much sir😊

[humaira]

Option c sir.
x/1-r=5
We get r=1-x/5
But to form an infinite Gp |r|<1
So -1<r<1 from here we get 0<x<10..
Thank you sir for these questions 😊

[rajeshsagar]

Option C i.e 0<x<10
Because in infinite GP
the 1 > r > -1

[shivanshprajapati]

sir n ki value 1 au 2 put karke saare options eliminate bhi kar sakte hai

[jatin]

Thank you sir revise ho jaata h daalte raha karo

[creativepassionhobbies]

Option C sir.
Let common ratio be r
x / (1-r) = 5
x = 5 - 5r
For an infinite gp to not have sum as infinity, |r|<1
So, substituting -1, 1
0<x<10

[ShubhAgrawal-cq]

Option c sir the concept used | r | < 1

[IIT_JEE-v]

Sir eaise questions ka solution dimaag mein kaise aate hai?

[syed]

I found the general term of the sequence in numerator in the first question to solve it

[JEEIIT]

For time saving to do more Qs in adv
In this type of Qs(asked till n terms )
Put n=1, 2 and find out the ans

[SatyamChaurasiya-sv]

Sir really jee adv few questions very easy

[shobabarath]

Please clarify my doubt sir.Sir for the first question I took the sum to be S and divided it by 2 and subtracted from s. This form a series of This is not providing the answer why ?

[IITian_jee_mains_questions]

Option C sahi hoga sir ji
5=x/1-r , 1-r=x/5 , r=5-x/5
If |r|<1 then
-1<5-x/5<1 , , -5<5-x<5
-10<-x< 0
Negative se into karne per inequality change ho jayega so
0<x<10 is right 👍 answer
Option C ❤🎉🎉

[meghajain]

Answer is option C:

S=a/1-r
5=x/1-r
r=1-x/5
So we can say that -1<1-x/5<1 so x range comes out to be 0<x<10.

[shreeharikallapur]

Please pin 📍 sir
Conic Sections please 🙏 🙏 🙏 🙏 🥺 🥺 🥺 🥺 🥺 🥺 🥺🥺🥺🥺🥺🥺🥺🥺

[LowCastImpulse]

Itne easy questions nhi aate guys advance me chill