Let's assume that we have an arithmetic series here (I don't know if we do, or not)
An arithmetic series is given by
an+1 = an + (n-1)d
Where an+1 is the term following a1.....n represents the nth term of the series, and d is the common difference between successive terms
Let's call the first term a1. And the second term(a2) is just a1 + (2-1)d = a1 + d
So the sum of the first two terms is just
a1+ (a1 + d) = 5 or 2a1 + d = 5 .....call this Equation 1
And the third term (a3) is just .... a1 + (3-1)d = a1 + 2d
And the fourth term(a4) is just..... a1 + (4-1)d = a1 + 3d
And the sum of these two terms = 12...... so
a1 + 2d + a1 + 3d = 12 or 2a1 + 5d = 12 ...... call this Equation 2
Now multiplying Equation 1 through by -1 on both sides and adding it to Equation 2, we get
4d = 7 so d = 7/4
And substituting for d in Equation 1, we have
2a1 + 7/4 = 5
2a1 = 13/4
a1 = 13/8 and that's the first term.....Now, lets check our results
a2 = a1 + (2-1)(7/4) = 13/8 +7/4 = 27/8
So a1 + a2= 13/8 + 27/8 = 40/8 = 5 .....and that's OK
And the sum of a3 and a4 is given by Equation 2 = 2a1 + 5d = 12
So we have
2(13/8) + 5(7/4) =
26/8 + 35/4 =
26/8 + 70/8 =
96/8 = 12 .....and that's OK, too
So our series (so far) is just
13/8, 27/8, 41/8, 55/8
And the sum of the 10th and 13th terms is
[a1 + (9)d] + [a1 + (12)d] = 2a1 + 21d = 2(13/8) + 21(7/4) = 26/8 + 294/8 = 320/8 = 40
