# Find the partial sum

• Sep 28th 2009, 09:24 PM
flexus
Find the partial sum
find the partial sum

100
6n
n=10

ok this is what i did but i didnt get the same answer as the book can someone tell me what i did wrong please?

100
6n
n=10

60+66+72+78+...

a1=60 d=66-60--> d=6

c=a1-d --> c=60-6 --> c= 54

an= dn+c --> an= 6n+54 --> a100=6(100)+54 --> a100= 654

Sn= n/2(a1+an) --> Sn=100/2(60+654) -->

i looked in the back of the book and the correct answer is 30,030. what did i do wrong?
• Sep 28th 2009, 09:27 PM
mr fantastic
Quote:

Originally Posted by flexus
find the partial sum

100
6n
n=10

ok this is what i did but i didnt get the same answer as the book can someone tell me what i did wrong please?

100
6n
n=10

60+66+72+78+...

a1=60 d=66-60--> d=6

c=a1-d --> c=60-6 --> c= 54

an= dn+c --> an= 6n+54 --> a100=6(100)+54 --> a100= 654

Sn= n/2(a1+an) --> Sn=100/2(60+654) -->

i looked in the back of the book and the correct answer is 30,030. what did i do wrong?

Arithmetic series: a = 60, d = 6, number of terms = 91. Substitute into the usual formula.
• Sep 28th 2009, 09:28 PM
tanujkush
sum of natural numbers from 1 to n is n(n+1)/2

= 6*{(sum from n=1 to n=100) - (sum from n=1 to n=9)}
= 6{100*101/2 - 9*10/2}
= 6{5050 - 45}
= 6{5005}
=30030
• Sep 28th 2009, 09:43 PM
flexus
Quote:

Originally Posted by mr fantastic
Arithmetic series: a = 60, d = 6, number of terms = 91. Substitute into the usual formula.

how did you get 91 for number of terms?
• Sep 28th 2009, 09:45 PM
mr fantastic
Quote:

Originally Posted by flexus
how did you get 91 for number of terms?

n = 10 to n = 100. How many values of n is that ....?
• Sep 28th 2009, 09:48 PM
flexus
Quote:

Originally Posted by mr fantastic
n = 10 to n = 100. How many values of n is that ....?

100 -10 = 90 right? how did you get the 1
• Sep 28th 2009, 09:53 PM
mr fantastic
Quote:

Originally Posted by flexus
100 -10 = 90 right? how did you get the 1

There are 90 numbers from 11 to 100 inclusive. Therefore from 10 to 100 there are 91. Maybe you ought to write them all out and count them all.