```Question 600581
The nth term of an arithmetic sequence is given by the formula a+(n-1)d and the sum of the first n terms of an arithmetic sequence is given by the formula.

n/2 * (2a+(n-1)d) where a is the first term and d is the common difference

Substituting 1 for a and -2 for d in the first equation the nth term of the given sequence works out to 1+(n-1)(-2), i.e. 3-2n

Since -13 is the term up to which sum of terms is required we use the equation 3-2n=-13 to give us the value of n for which the term is of value -13.
3-2n=-13 implies n=8 which means that n=8, i.e. -13 is the 8th term in the sequence.

Using the same values of a and d and n=8 in the formula for sum of n terms we end up with

8/2 * (2+(8-1)(-2)) = 8 * (1-8+1) = 8 * (-6) = -48 as the required sum.

(also note sum of n terms for this particular sequence is given by the formula n/2 (2+(n-1)(-2))
i.e. n((1-(n-1))
i.e. n(1-n+1)
i.e. n(2-n). Substituting 8 for n gives the answer as -48)

In the sequence -17, a2, a3, 1 a=-17 4th term= a+(4-1)d= -17 + 3d = 1 which implies d = 6. therefore a2 = -17 + 6 = -11 and a3 = a2 + 6 = -11 + 6 = -5.

Similarly in the sequence 1, a2, a3, a4, -35 a=1 5th term = a + (5-1)d = 1 + 4d = -35 which implies d=-9 and therefore a2 = 1-9 =-8, a3 = -8-9 = -17 and a4 = -17-9 = -26.

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