Aptitude - Numbers - Discussion

How you get 3n(n+1) = 1800?

Solution is;
n/2 [2 x 6 + (n - 1) x 6] = 1800,
n/2 [12 + (6n-6)] = 1800,
12n/2 + 6n2/2 - 6n/2=1800,
12n+6n2-6n/2=1800,
6(2n+1n2+1n)/2=1800,
6(1n2+1n)/2=1800,
3(1n2+1n)=1800,
3n(n+1)=1800.

1800 = n/2[2a+(n-1)d],
1800 = n/2[2*6+(n-1)6],
3600 = n[12+6n-6],
3600 = n[6n+6],
6n^2+6n-3600 = 0,
6(n^2+n-600) = 0,
n^2+n-600 = 0,
n^2+25n-24n-600 = 0,
n(n+25)-24(n+25) = 0,
(n+25)(n-24) = 0,
n=-25 , n = 24,
Negative value is not valid.
Therefore, n=24.

@All.

This method will clear your doubt.

6+12+18...... = 1800
6(1+2+3+....) = 1800
1+2+3 = 300.
so, n(n+1)/2 = 300 (this is the formula to calculate the sum of the natural number since 1+2+3+....series is a natural number series.

n(n+1) = 600.
Now think what could be n.
24^576
25^625 since(n+1)

Now,
n(n+1) = n^2+n,
take n = 24^2,
so 24^2 + 25 = 600,
25 = 600 - 24^2 = 24 so the answer is 24.

1) tn = a+(n-1)d --> [ a=first term, n=no. of terms, d=difference between terms, tn=last term ]
tn = 6+(n-1)6
tn = 6+6n-6
tn = 6n

2) sn = n/2(a+l) -----------------> [ sn=sum, n=no. of terms, a=first term, l=last term ]
1800 = n/2(6+tn) ------------> [ l=tn from (1) ]
3600 = n(6+6n),
600 = n(1+n),
600 = n+n^2,
n^2+n-600 = 0.
n^2+25n-24n-600 = 0,
n(n+25)-24(n+25) = 0,
(n+25)(n-24) = 0,
n=-25 or n=24.
n=24 ---------> [ consider positive value].

So, the answer is 24.

Nice move @Albin.

Please explain me 3rd step.

How you get 3n(n+1)=1800?

Please explain me the 3rd step.

Why let 25n and 24n? Please explain me.

Please explain, How come 25 and 24?

Please tell me why we have taken this formula, i.e n/2[2a+(n-1)]d.

In the previous problem of (2+4+6+...30) we have taken the formula as [a+(n-1)d]. So tell me which formula must be used for which problem.