The sum of the first n terms of an arithmetic series whose nth term is

Question

The sum of the first n terms of an arithmetic series whose nth term is n can be calculated by the formula \(\frac{n(n + 1)}{2}\). Which one of the following equals the sum of the first eight terms in a series whose nth term is 2n ?

(A) 24
(B) 48
(C) 56
(D) 72
(E) 96

(A) 24
(B) 48
(C) 56
(D) 72
(E) 96

Archit3110 · Answer

given
 \frac{n(n + 1)}{2} 
which can be written as
8*9/2 = 36 = n
so 2n = 36*2 ; 72
IMO D

jfranciscocuencag · Answer

What I understood from this one is that they are asking for the sum of the first n terms multiplied by 2.

\frac{8(8+1)}{2} = 36

36*2 = 72

But..  Archit3110 , how did you get n^2+n?

I tried by substituting 2n instead of n:

\frac{2n(2n+1)}{2}

n(2n+1)

2n^2+n

nick1816 · Answer

2+4+6+8+10+12+14+16
=2(1+2+3+4+5+6+7+8)
=2*(8*9)/2
=72

Archit3110 · Answer

jfranciscocuencag i did some typo error... edited thanks

MathRevolution · Answer

The sum of the first n terms of an arithmetic series whose nth term is n:  \frac{  }{ 2}

The sum of the first n terms of an arithmetic series whose nth term is 2n: 2 *  \frac{  }{ 2}  = n * (n + 1)

The sum of the first 8 terms of an arithmetic series whose nth term is 2n: 8 * (8 + 1) = 8 * 9 = 72

Answer D

shubhamjha22 · Answer

The question is basically asking the sum of 1st 8 even terms:

n(n+1)/2 is the sum of consecutive integers. but the nth term is always 2n or even

for n=8 nth term in the series is 16 
therefore Sn= 2+4+6+8+10+12+14+16= 72. Therefore D. (I took the sum from 0, for some reason and screwed up the answer )

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The sum of the first n terms of an arithmetic series whose nth term is

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The sum of the first n terms of an arithmetic series whose nth term is

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