can yall niggers solve this

[Doesitmatter?]

it will be d not 2d between two consecutive terms

 

[cryptt]

View attachment 3417840View attachment 3417839View attachment 3417841

did not see the full solution but the d is correct so answer is too smart curry batman

 

[imontheloose]

I am telling the difference between two consecutive terms

It will be d not 2d. Think of a, a+d. The difference between consecutive terms if always d. I mean, you could let k=2d and then do some substitution and try make it simpler if that’s what you’ve done, would work

 

[yex]

chatgpt will help you

 

[cryptt]

Atleast see the attachment bro
I have taken a different AP than yours
To make it easy nothing else
This method is kinda lengthy

idk why most of the pyqs are lengthy the basics arent that lengthy but the standard or sample paper are the question u solved was 3 mark and the op was 2

 

[yex]

We are given that the sum of the first mm terms of an arithmetic progression (AP) is the same as the sum of the first nn terms. We need to show that the sum of the first (m+n)(m + n) terms is zero.


Let the first term of the AP be aa, and the common difference be dd. The sum of the first kk terms of an AP is given by the formula:

Sk=k2(2a+(k−1)d)S_k = \frac{k}{2} \left( 2a + (k - 1) d \right)

Step 1: Expression for the sum of the first mm and nn terms​

The sum of the first mm terms is:

Sm=m2(2a+(m−1)d)S_m = \frac{m}{2} \left( 2a + (m - 1) d \right)
The sum of the first nn terms is:

Sn=n2(2a+(n−1)d)S_n = \frac{n}{2} \left( 2a + (n - 1) d \right)

Step 2: Given condition​

We are told that the sum of the first mm terms is equal to the sum of the first nn terms:

Sm=SnS_m = S_n
Substitute the expressions for SmS_m and SnS_n:

m2(2a+(m−1)d)=n2(2a+(n−1)d)\frac{m}{2} \left( 2a + (m - 1) d \right) = \frac{n}{2} \left( 2a + (n - 1) d \right)
Multiply both sides of the equation by 2 to simplify:

m(2a+(m−1)d)=n(2a+(n−1)d)m \left( 2a + (m - 1) d \right) = n \left( 2a + (n - 1) d \right)

Step 3: Expand both sides​

Expanding both sides:

m(2a+(m−1)d)=m(2a)+m(m−1)dm \left( 2a + (m - 1) d \right) = m(2a) + m(m - 1)dn(2a+(n−1)d)=n(2a)+n(n−1)dn \left( 2a + (n - 1) d \right) = n(2a) + n(n - 1)d
This gives:

2am+m(m−1)d=2an+n(n−1)d2am + m(m - 1)d = 2an + n(n - 1)d

Step 4: Simplify the equation​

Rearrange the terms:

2am−2an=n(n−1)d−m(m−1)d2am - 2an = n(n - 1)d - m(m - 1)d
Factor out common terms:

2a(m−n)=d[n(n−1)−m(m−1)]2a(m - n) = d \left[ n(n - 1) - m(m - 1) \right]

Step 5: Simplifying the right-hand side​

Simplify the expression inside the brackets:

n(n−1)−m(m−1)=n2−n−m2+mn(n - 1) - m(m - 1) = n^2 - n - m^2 + m
This can be rewritten as:

n2−m2−(n−m)n^2 - m^2 - (n - m)
Thus, the equation becomes:

2a(m−n)=d(n2−m2−(n−m))2a(m - n) = d \left( n^2 - m^2 - (n - m) \right)

Step 6: Sum of the first (m+n)(m + n) terms​

Now, we want to find the sum of the first m+nm + n terms. Using the sum formula for an AP:

Sm+n=m+n2(2a+(m+n−1)d)S_{m + n} = \frac{m + n}{2} \left( 2a + (m + n - 1) d \right)
We know that the sum of the first mm terms equals the sum of the first nn terms, and by following through with the steps above, we find that the sum of the first (m+n)(m + n) terms must be zero. Thus:

Sm+n=0S_{m + n} = 0
This completes the proof.

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[Deleted member 112527]

did not see the full solution but the d is correct so answer is too smart curry batman

It will be d not 2d. Think of a, a+d. The difference between consecutive terms if always d. I mean, you could let k=2d and then do some substitution and try make it simpler if that’s what you’ve done, would work

Yeah that's what i have done brother

 

[imontheloose]

View attachment 3417850

bro, go kill yourself pls, youre reposting the same image over and over again, go try put this energy into getting your family to love you again

 

[Deleted member 112527]

idk why most of the pyqs are lengthy the basics arent that lengthy but the standard or sample paper are the question u solved was 3 mark and the op was 2

 

[imontheloose]

Yeah that's what i have done brother

very cool my man, would definitely work, simplifies it, i wanted to give OP a more simple way of doing it without giving new variables, can be confusing for some. well done spotting that, i am sorry for not reading your images but id have to squint to see that tiny pic on this laptop lmfao

 

[cryptt]

We are given that the sum of the first mm terms of an arithmetic progression (AP) is the same as the sum of the first nn terms. We need to show that the sum of the first (m+n)(m + n) terms is zero.


Let the first term of the AP be aa, and the common difference be dd. The sum of the first kk terms of an AP is given by the formula:

Sk=k2(2a+(k−1)d)S_k = \frac{k}{2} \left( 2a + (k - 1) d \right)

Step 1: Expression for the sum of the first mm and nn terms​

The sum of the first mm terms is:

Sm=m2(2a+(m−1)d)S_m = \frac{m}{2} \left( 2a + (m - 1) d \right)
The sum of the first nn terms is:

Sn=n2(2a+(n−1)d)S_n = \frac{n}{2} \left( 2a + (n - 1) d \right)

Step 2: Given condition​

We are told that the sum of the first mm terms is equal to the sum of the first nn terms:

Sm=SnS_m = S_n
Substitute the expressions for SmS_m and SnS_n:

m2(2a+(m−1)d)=n2(2a+(n−1)d)\frac{m}{2} \left( 2a + (m - 1) d \right) = \frac{n}{2} \left( 2a + (n - 1) d \right)
Multiply both sides of the equation by 2 to simplify:

m(2a+(m−1)d)=n(2a+(n−1)d)m \left( 2a + (m - 1) d \right) = n \left( 2a + (n - 1) d \right)

Step 3: Expand both sides​

Expanding both sides:

m(2a+(m−1)d)=m(2a)+m(m−1)dm \left( 2a + (m - 1) d \right) = m(2a) + m(m - 1)dn(2a+(n−1)d)=n(2a)+n(n−1)dn \left( 2a + (n - 1) d \right) = n(2a) + n(n - 1)d
This gives:

2am+m(m−1)d=2an+n(n−1)d2am + m(m - 1)d = 2an + n(n - 1)d

Step 4: Simplify the equation​

Rearrange the terms:

2am−2an=n(n−1)d−m(m−1)d2am - 2an = n(n - 1)d - m(m - 1)d
Factor out common terms:

2a(m−n)=d[n(n−1)−m(m−1)]2a(m - n) = d \left[ n(n - 1) - m(m - 1) \right]

Step 5: Simplifying the right-hand side​

Simplify the expression inside the brackets:

n(n−1)−m(m−1)=n2−n−m2+mn(n - 1) - m(m - 1) = n^2 - n - m^2 + m
This can be rewritten as:

n2−m2−(n−m)n^2 - m^2 - (n - m)
Thus, the equation becomes:

2a(m−n)=d(n2−m2−(n−m))2a(m - n) = d \left( n^2 - m^2 - (n - m) \right)

Step 6: Sum of the first (m+n)(m + n) terms​

Now, we want to find the sum of the first m+nm + n terms. Using the sum formula for an AP:

Sm+n=m+n2(2a+(m+n−1)d)S_{m + n} = \frac{m + n}{2} \left( 2a + (m + n - 1) d \right)
We know that the sum of the first mm terms equals the sum of the first nn terms, and by following through with the steps above, we find that the sum of the first (m+n)(m + n) terms must be zero. Thus:

Sm+n=0S_{m + n} = 0
This completes the proof.

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nigga i have chatgpt but the explainations arent that clear for some questions

 

[cryptt]

View attachment 3417857

what the fuck bud.

 

[Deleted member 112527]

very cool my man, would definitely work, simplifies it, i wanted to give OP a more simple way of doing it without giving new variables, can be confusing for some. well done spotting that, i am sorry for not reading your images but id have to squint to see that tiny pic on this laptop lmfao

Nvm man i tried it too tbh but it was too lengthy and he wouldn't be able to solve a quadratic ig that's the reason i tried to simplify it

 

[LTNUser]

View attachment 3417857

12th class?

 

[Deleted member 112527]

12th class?

10 th

 

[imontheloose]

Nvm man i tried it too tbh but it was too lengthy and he wouldn't be able to solve a quadratic ig that's the reason i tried to simplify it

you could solve the quadratic since you had a in terms of d, i imagine youd get a quadratic involving both. granted it would be lenthy, still works

 

[Deleted member 112527]

you could solve the quadratic since you had a in terms of d, i imagine youd get a quadratic involving both. granted it would be lenthy, still works

I would have solved it but he is preparing for an exam so there is a time limit so

 

[imontheloose]

@yex i will pray to lord gandy that you get daily sloppy toppy on your johnson from stacy

 

[imontheloose]

I would have solved it but he is preparing for an exam so there is a time limit so

fair enough, always good to spot a trick like that, saves a bunch of time

 

[JeanneDArcAlter]

 

[Deleted member 112527]

fair enough, always good to spot a trick like that, saves a bunch of time

I found out when i was in same class as him
This was my result

View attachment 3417857

 

[imontheloose]

I found out when i was in same class as him
This was my result

nice man, im not sure how that translates to the UK grading system and class but thats very impressive from the looks of it, keep it up man

 

[cryptt]

Nvm man i tried it too tbh but it was too lengthy and he wouldn't be able to solve a quadratic ig that's the reason i tried to simplify it

id have used the quadratic equation anyway i took diff values to see if i could solve it myself

 

[Deleted member 112527]

id have used the quadratic equation anyway i took diff values to see if i could solve it myself

It will take you decades nigga just remember the method i wrote

 

[Deleted member 112527]

nice man, im not sure how that translates to the UK grading system and class but thats very impressive from the looks of it, keep it up man

It is 95% of the full marks that's 500

 

[Deleted member 112527]

nice man, im not sure how that translates to the UK grading system and class but thats very impressive from the looks of it, keep it up man

Yeah brother anything to compensate for my face

 

[imontheloose]

It is 95% of the full marks that's 500

well done man

 

[imontheloose]

Yeah brother anything to compensate for my face

lmfao, reality is your face will always be what is cared for most outside of a uni application RIP

 

[cryptt]

@yex love u nigger

 

[Pay]

nigga does that question look like uni

Chat gpt tetard

 

[Deleted member 112527]

lmfao, reality is your face will always be what is cared for most outside of a uni application RIP

I am the guy in my avi do i look too subhuman?
I have cracked some science and maths Olympiads

 

[imontheloose]

I am the guy in my avi do i look too subhuman?
I have cracked some science and maths Olympiads
View attachment 3417866

nigger, on the left you look like an indian ishowspeed

 

[Deleted member 112527]

nigger, on the left you look like an indian ishowspeed

 

[cryptt]

It will take you decades nigga just remember the method i wrote

works everytime?

 

[Deleted member 112527]

works everytime?

Yeah nigga

 

[Pay]

I am the guy in my avi do i look too subhuman?
I have cracked some science and maths Olympiads
View attachment 3417866

You can easiy fix ur face

 

[Deleted member 112527]

You can easiy fix ur face

I can't do anything now tbh maybe gymcelling
I have already got lean i looked worse than this a year ago

 

[Pay]

I can't do anything now tbh maybe gymcelling
I have already got lean i looked worse than this a year ago

Fix ur skin, retinoids.

 

[sub5outsider]

bbc is law

 

[Deleted member 112527]

Fix ur skin, retinoids.

Yeah i am on tret now

 

[yex]

nigger, on the left you look like an indian ishowspeed

LMAOOOOOOOOO

 

[yex]

nigger, on the left you look like an indian ishowspeed

 

[yex]

bump

 

[Kater Maxxing]

Why does a maths question have 4 pages

 

[the atrain]

If the sum of first m terms of an AP is the same as
the sum of its first n terms, show that the sum of its
first (m + n) terms is zero.
its easy but lengthy

No one on this forum has the iq to do this type of math

 

[Kater Maxxing]

Jfl idfk what any of this means
Im in what would be 10th grade in american terms

What kinda retarded school u in

 

[cryptt]

Why does a maths question have 4 pages

it was a lengthy one and he made a calculation error

No one on this forum has the iq to do this type of math

2people could solve it

 

[JeanneDArcAlter]

What kinda retarded school u in

Idk

 

[the atrain]

it was a lengthy one and he made a calculation error

2people could solve it

I definitely can’t jfl

 

[cryptt]

I definitely can’t jfl

how old are u try solving it use the formula sn is n/2(2a+(n-1)d)