Question 431863
1. Looking at the terms of the series, we see that the n-th term is one less than n^2.  For example, the 3rd term = 8 = 3^2 - 1
So, we can write the series as:
{{{sum(i^2-1,i=1,7)}}}
2. The arithmetic sequence 3/4,13/12,17/12,... has a common difference of 1/3.
So we can write: 
a_n = a_1 + (1/3)(n-1)
Where a_1 is the first term of the sequence
So the formula for the series becomes:
a_n = 3/4 + n/3 - 1/3 -> a_n = 5/12 + n/3
So the 10th term in the sequence is a_10 = 5/12 + 10/3 = 5/12 + 40/12 = 45/12
3. The common difference is a-(a-b) = b
So the formula for the series is:
(a-b) + (n-1)b
Putting in n=6 gives:
(a-b) + 5b = a + 4b
4. An arithmetic series is written a_n = a_1 + (n-1)d
So we have a_n = a_1 + 3(n-1)
For n=10, we have
23 = a_1 + 3(9), or 23 = a_1 + 27
Therefore a_1 = -4