Question 856145
<pre>
You can also use the other formula:

{{{S[n]=expr(n/2)(2a[1]+(n-1)d)}}}

8, 14, 20 …, if there are 22 terms

{{{a[1]}}} = the first term which = 8

d = common difference = 2nd term - 1st term = 14 - 8 = 6

check d:

d = common difference = 3rd term - 2nd term = 20 - 14 = 6

So d = 6

n = the number of terms = 22

Substitute {{{a[1]=8}}}, {{{d=6}}}, and {{{n=22}}}

{{{S[n]=(n/2)(2a[1]+(n-1)d)}}}

{{{S[22]=(22/2)(2(8)+(22-1)6)}}}

{{{S[22]=(11)(16+(21)6)}}}

{{{S[22]=(11)(16+126)}}}

{{{S[22]=(11)(142)}}}


{{{S[22]=1562}}}




Edwin</pre>