# Sum 7+10+13+15+...+97+100 = ?

Arithmetic and Geometric progressions.

### Sum 7+10+13+15+...+97+100 = ?

Find the sum:

$P= 7+10+13+15+...+97+100$

dduclam

Posts: 36
Joined: Sat Dec 29, 2007 10:42 am
Location: HUCE-Vietnam

Are you shure that the problem is not
P = 7 + 10 + 13 + 16 + ... + 97 + 100 ?
Math Tutor

Posts: 313
Joined: Sun Oct 09, 2005 11:37 am

### edit

teacher wrote:Are you shure that the problem is not
P = 7 + 10 + 13 + 16 + ... + 97 + 100 ?

Yes,thank you,teacher P = 7 + 10 + 13 + 16 + ... + 97 + 100
math is my life
dduclam

Posts: 36
Joined: Sat Dec 29, 2007 10:42 am
Location: HUCE-Vietnam

This is an arithmetical progression.
icb

Posts: 2
Joined: Wed May 14, 2008 7:55 am

1st term is 7, last term is 100, d- common difference is 3.

So we need only n, to find out the Sn=1/2 (a 1st term + a nth term) * n

=1/2*(7+100)*n

100=(7+31*3) , where a=7, 3=d, and a(nth) term = 100, and 31 is the (n-1)th term.

total terms= 32.

Sn= 1/2(7+100)*32= 16*107=1712.
Prasanna Chandran
prasanna_chandran

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