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

[dduclam]

Find the sum:

[tex]P= 7+10+13+15+...+97+100[/tex]

[Math Tutor]

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

[dduclam]

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

[icb]

This is an arithmetical progression.

[prasanna_chandran]

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.

[burgess]

Find the sum:

[tex]P= 7+10+13+16+...+97+100[/tex]

It is an arithmetic progression
a=7, n,d=3
first we will find out number of terms
7 + (n-1)*3 =100
n-1=93/3=31 => n=32

Sum = n/2 *[first term + last term] = 16*107

[Alicelewis11]

Yes, we can use this formula to sum all integers (N(N + 1))/2...............

[burgess]

Use the arithmetic series formulas

Find the sum:

[tex]P= 7+10+13+15+...+97+100[/tex]

[zaiveca]

This is really great!

[leesajohnson]

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

The answer for this question is 1712.