Question 978276
<pre>[You did not give enough information.  You had this:</pre>an=35 and s14=931 a sub n ,common diffence and a sub 1<pre>However I guessed that a<sub>4</sub>=35 was given, and rewrote your question as above.]

We use the nth term formula:

{{{a[n]}}}{{{""=""}}}{{{a[1]+(n-1)d}}}

Substitute n=4, then a<sub>4</sub>=35

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

{{{35}}}{{{""=""}}}{{{a[1]+3d}}}

We use the sum formula:

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

Substitute n=14, then S<sub>14</sub>=931

{{{S[14]}}}{{{""=""}}}{{{expr(14/2)(2a[1]+(14-1)d^"")}}}

{{{931}}}{{{""=""}}}{{{7(2a[1]+13d^"")}}}

Both sides can be divided through by 7:

{{{133}}}{{{""=""}}}{{{2a[1]+13d^"")}}}

So now we have this system of equations:

{{{system(35=a[1]+3d,133=2a[1]+13d)}}}

Solve that and get a<sub>1</sub>=8, d=8.

If my guess of a<sub>4</sub>=35 is correct, then your sequence was

8, 17, 26, 35, 44, 53, 62, 71, 80, 89, 98, 107, 116, 125 

That sequence of 14 terms does have sum 931.

If I guessed wrong, then you can use the above as a guide.

Edwin</pre>