I mentioned the calculus-based justification for that. A rigorous non-calculus based proof would be more painful; please take my word for it that it will be of this form if you don’t like calculus 🙂
sir, as you mentioned it can be solved thru calculus, so I tried solving it in two ways, one gave the correct answer, other didn’t, can you please explain why this happened?
1) WRONG
T2 = 4 = 2 ax + b (first derivative) (keeping n=2),
we get 4=4+b;
so, b=0
2) RIGHT
taking second order derivative of ax^2+bx+c = 0,
2a = 2 (second order difference is 2)
a = 1,
now solving the two equations keeping value of a in it.
t1 = a + b + c = 3 and t2= 4a + 2b + c = 5,
at a =1, becomes
b + c = 2
2b + c = 1
we get, b= -1 and c = 3
T2 is not the first derivative :P. T2 is the second term. Also your “a” seems to have vanished in this step
“4 = 2 ax + b (first derivative) (keeping n=2),
we get 4=4+b;”
but that is irrelevant as anyway you are applying it wrong to begin with. You don’t apply derivative at any random point of your choice.
You should instead have done
ax^2 +bx +c
First derivative 2ax + b (this could be used to find the x for max/min by putting it equal to 0)
Second derivative = 2a = second level difference which is 2 in this case.
CAT-holics 
Sir, can this question be solved with Calculus?
Sir, can this question be solved using Calculus?
Perhaps, but it seems too much effort (quite aside from being out of scope for CAT)? It is a discrete problem, not a continuous one..
regards
J
How did you arrive at that expression an^2+bn+c for the nth term?
please explain, Thanks
I mentioned the calculus-based justification for that. A rigorous non-calculus based proof would be more painful; please take my word for it that it will be of this form if you don’t like calculus 🙂
regards
J
sir, as you mentioned it can be solved thru calculus, so I tried solving it in two ways, one gave the correct answer, other didn’t, can you please explain why this happened?
1) WRONG
T2 = 4 = 2 ax + b (first derivative) (keeping n=2),
we get 4=4+b;
so, b=0
2) RIGHT
taking second order derivative of ax^2+bx+c = 0,
2a = 2 (second order difference is 2)
a = 1,
now solving the two equations keeping value of a in it.
t1 = a + b + c = 3 and t2= 4a + 2b + c = 5,
at a =1, becomes
b + c = 2
2b + c = 1
we get, b= -1 and c = 3
T2 is not the first derivative :P. T2 is the second term. Also your “a” seems to have vanished in this step
“4 = 2 ax + b (first derivative) (keeping n=2),
we get 4=4+b;”
but that is irrelevant as anyway you are applying it wrong to begin with. You don’t apply derivative at any random point of your choice.
You should instead have done
ax^2 +bx +c
First derivative 2ax + b (this could be used to find the x for max/min by putting it equal to 0)
Second derivative = 2a = second level difference which is 2 in this case.
regards
J