What is the sum of a 34-term arithmetic sequence where the first term is 14 and the last term is 146?

2,550
2,720
2,890
2,982

Question

What is the sum of a 34-term arithmetic sequence where the first term is 14 and the last term is 146?
 	
	2,550
	2,720
	2,890
	2,982

lgbasallote · Answer

use the equation $$\large S_n = \frac{n(A_1 + A_n)}{2}$$

lgbasallote · Answer

where Sn is the sum of the series
n = number of terms in the series
A1 = first term
An = last term

anonymous · Answer

2,720?

anonymous · Answer

Okay, this one's a little more abstract.  We have the common difference is
d = (a34 - a1)/33 = (146 - 14)/33 = 4
Now
a1 = a1
a2 = a1 + d
a3 = a1 + 2d 
...
a34 = a1 + 33d
So to add all these up we need to add the first 33 natural numbers together.
the formula is 1 + 2 + ... + n = n(n+1)/2
Which I'll explain if you'd like.
Using this formula we have that the sum from a1 to a34 is
34*a1 + (33(34)/2)d = 34*14 + 33*17*4 = 476 + 2244 = 2720

anonymous · Answer

Okay, well, there's a longwinded explanation of how it works

jim_thompson5910 · Answer

yes you have it right kerber123