how many terms of the progression 50,46,42,... must be taken for its sum to be equal to 0?

Let [tex]S_{n}[/tex] be the sum of sequence up to n terms. since it's a arithmetic sequence, the difference is constant, so
[tex]d=u_{2}-u_{1}\\=46-50\\=-4[/tex]
now using n-term formula, we get
[tex]u_{n}=u_{1}+(n-1)d\\=50+(n-1)\times(-4)\\=50-4n+4\\=54-4n[/tex]
lets use our logic, the sequence is (50,46,42,38,...10,6,2,-2,-6,-10,...,-42,-46,-50,...). the sequence is symmetry to 0
[tex]50+46+42+...+6+2+(-2)+(-6)+...+(-46)+(-50)=(50-50)+(46-46)+...(2-2)=0+0+...+0=0[/tex]
so, when [tex]u_{n}=-50[/tex], the sum is equal to 0. using n-term formula, we get the n
[tex]u_{n}=54-4n\\-50=54-4n\\4n=104\\n=26[/tex]

the other way is using direct formula. here is the derivative:
[tex]S_{n}=u_{1}+(u_{1}+d)+(u_{1}+2d)+...+(u_{1}+(n-1)d)\\=u_{1}\times n+(0\times d+1\times 1+2\times d+...+(n-1)\times d)\\=u_{1}\times n+(0+1+2+...+(n-1))\times d\\=u_{1}\times n+\frac{(n-1)((n-1)+1)}{2}\times d\\=u_{1}\times n+\frac{n(n-1)}{2}\times d\\=\frac{2u_{1}n+n(n-1)\times d}{2}\\=\frac{n(2u_{1}+(n-1)d)}{2}[/tex]

just follow the step below
now, using series formula for arithmetic series, we get
[tex]S_{n}=\frac{n(2u_{1}+(n-1)d)}{2}\\=\frac{n(2\times 50+(n-1)\times(-4))}{2}\\=\frac{n(100-4n+4)}{2}\\=\frac{n(104-4n)}{2}[/tex]

the last step, just substitute Sn by 0
[tex]S_{n}=\frac{n(104-4n)}{2}\\0=\frac{n(104-4n)}{2}\\n=0\vee 104-4n=0\\n_{1}=0\vee n_{2}=26[/tex]

Bab Rows and series

50,46,42

initial Value = 50 = IV
different = 46 - 50 = - 4 = D

Un = IV + (n - 1) D
Un = 50 + (n - 1) -4
Un = 50 + -4n + 4
Un = 54 -4n


Un = - 50

Un = 54 - 4n
-50 = 54 - 4n (Both segments are exchanged)
4n = 54 + 50
4n = 104
n = 104/4
n = 26


So,many developmental requirements = 26