Whats the sum of the first 50 terms?

$ \huge { 12 - -\frac{7}{10}n } $

I am getting \ ( \huge -\frac{445}{2} \) is that right because it is stating it is wrong. They stat it is $ \huge -\frac{585}{2} $

I am using $ \huge \frac{n}{2}[2*a_1 + (n-1)d) ] to solve$

Question

Whats the sum of the first 50 terms?

$  \huge { 12 - -\frac{7}{10}n  }  $

I am getting \ ( \huge -\frac{445}{2}  \) is that right because it is stating it is wrong. They stat it is $ \huge -\frac{585}{2} $

I am using $ \huge \frac{n}{2}[2*a_1 + (n-1)d) ] to solve$

anonymous · Answer

Whats the sum of the first 50 terms?

$  \huge { 12 - -\frac{7}{10}n  }  $

I am getting \ ( \huge -\frac{445}{2}  \) is that right because it is stating it is wrong. They stat it is $ \huge -\frac{585}{2} $

I am using $ \huge \frac{n}{2}[2*a_1 + (n-1)d) ] $ to solve

anonymous · Answer

$ \huge \frac{50}{2}[2*\frac{127}{10} + (50-1)*-\frac{7}{10} ]$

anonymous · Answer

This $ \huge \huge { 12 - -\frac{7}{10}n  }$ SHOULD BE  $  \huge\huge { 12 - \frac{7}{10}n  } $

anonymous · Answer

Wait, so is the sequence $a_n=12-\frac{7}{10}n$,     $n>0$?

anonymous · Answer

Yes

anonymous · Answer

Ok, if so, I think what you did wrong is that your first term is wrong.$$a_1=12-\frac{7}{10}=\frac{113}{10}$$

anonymous · Answer

Ah and I got 127/10. Now how did I get 127......  That was the problem.... I hate mis calculating problems lol Thank you. It came up to be correct.

anonymous · Answer

lol np. Everyone makes these kinds of mistakes from time to time. :)