Question 257426
The first three terms of an arithmetic sequence, in order, are 2x + 4, 5x – 4, and
3x + 4. What is the sum of the first ten terms of the sequence?

<pre><font size = 4 color = "indigo"><b>
If it's an arithmetic sequence then

the second term minus the first term must be equal to third term minus the second term.

So 

(5x-4) - (2x+4) = (3x+4) - (5x-4)

5x - 4 - 2x - 4 = 3x + 4 - 5x + 4
 
         3x - 8 = -2x + 8
    
        3x + 2x = 8 + 8

             5x = 16

              x = {{{16/5}}}

     
2x + 4, 5x – 4, and 3x + 4

Substitute {{{16/5}}} for x in all three terms:

{{{2x + 4}}} 
{{{2(16/5)+4)}}}
{{{32/5+20/5}}} 
{{{52/5}}}


{{{5x – 4}}}
{{{5(16/5)-4}}}
{{{16-4}}}
{{{12}}}

{{{3x + 4}}}
{{{3(16/5) + 4}}}
{{{48/5+4}}}
{{{48/5+20/5}}}
{{{68/5}}}

So the common difference d is 

{{{12 - 52/5}}}
{{{60/5-52/5}}}
{{{8/5}}}

the sum of the first n terms is given by the formula

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

So we substitute {{{n=10}}} and {{{d=8/5}}} and {{{a[1]=52/5}}}

{{{ S[10]=(10/2)(2(52/5)+((10)-1)(8/5))}}}

{{{S[10]=(5)(104/5+(9)(8/5))}}}

{{{S[10]=(5)(104/5+72/5))}}}

{{{S[10]=(5)(176/5))}}}

{{{S[10]=176}}}

Edwin</pre>