Question 484324
In the arithmetic sequence, the 3rd term is 4 and the 8th term is 49. Find the 1st term, the common difference and the sum of the first ten terms.
<pre>
{{{a[n]=a[1]+(n-1)d}}}

>>...the 3rd term is 4...<<

{{{a[3]=a[1]+(3-1)d}}}
{{{4=a[1]+2d}}}
{{{a[1]+2d=4}}}


>>...the 8th term is 49...<<

{{{a[8]=a[1]+(8-1)d}}}
{{{49=a[1]+7d}}}
{{{a[1]+7d=49}}}


So we have a system of equations:

{{{system(a[1]+2d=4,a[1]+7d=49)}}}

Solve by substitution or elimination and get

{{{a[1]=-14}}} and {{{d=9}}}

first term is -14,

Here is the sequence to 10 terms:

-14, -5, 4, 13, 22, 31, 40, 49, 58, 67 

To get the sum of the first 10 terms, 
we use the formula:

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

{{{S[10] = expr(10/2)(2(-14)+(10-1)9)}}}

{{{S[10] = 5(-28+(9)9)}}}

{{{S[10] = 5(-28+81)}}}

{{{S[10] = 5(53)}}}

{{{S[10] = 265}}}

sum =  265

Edwin</pre>