Find the sum of the first 15 terms of the series whose nth

Question

Find the sum of the first 15 terms of the series whose nth term is (4n+1).

A. 485
B. 495
C. 505
D. 630

mau5 · Accepted Answer

First term : 4*1+1 = 5.

15th term : 4*15+1= 61

Sum of 15 terms :  \frac{No of terms*(first term + last term)}{2} = \frac{15*(61+5)}{2} = 495.

B.

jrymbei · Answer

Thanks for the explanation...it was so simple!!!

saurabhprashar · Answer

Formula used: Sum of n terms = Average (First and Last term) * Number of terms.

First term : n=1, (4*1+1) = 5
Last term : n=15, (4*15 + 1) = 61

Sum = (5+61)/2 * 15 = 495. Answer B.

sisorayi01 · Answer

Its probably obvious.. but can someone explain why the formula for a arithmetic series is used here ... Sum of series n = (n/2)(A1 - An) ... as opposed to the formula for a geometric series ... Sum of series n =  /(1-r) where r is the common ratio and n is the nth term.

I understand the solution but am confused by the formula thinking that An=4n+1 is a geometric sequence given that you have to multiply by 4 to get the next term in the sequence. Appreciate your reply!!

THANKS!

Bunuel · Answer

The sequence is define by  a_n=4n+1 , thus:

a_1=4*1+1=5 ;
 a_2=4*2+1=9 ;
 a_3=4*3+1=13 ;
 a_4=4*4+1=17 ;
...

As you can see the sequence we have (5, 9, 13, 17, ...) is an arithmetic progression, not geometric.

shreyast · Answer

Adding to what Bunuel just posted, a geometric series would look like following
An = 4^n
And the series would look something like 4, 16, 64, 256, 1024, ...

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Find the sum of the first 15 terms of the series whose nth

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Find the sum of the first 15 terms of the series whose nth

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