Question: finding sum of first 22 terms in an arithmetic sequence when 3rd term is 9 and 7th term is 31.
There is one difference between any two successive terms and there are 4 differences between the 3rd term and the 7th terrm.
Using d for the difference between terms, 4d = 31 - 9 = 22.
Therefore, d = 5.5
The sum of the 1st n terms, Sn, is n times the average value, where A_av = (A1 + An)/2
And An = A1 + (n-1)d (i.e. the 1st term plus (n-1) differences)
So, A1 + An = A1 + A1 + (n-1)d
And, A_av = (2A1 + (n-1)d)/2
Then Sn = n(2*A1 + (n-1)d)/2
And S22 = 22(-4 + (21)*5.5)/2 = 11(-4 + 115.5) = 11*111.5 = 1226.5
Answer: S22 = 1226.5