other sequence

Let a and d be the first term and the common difference of the arithmetic sequence.
Sum of first 10 terms = (10/2)(2a+ 9d) = 50
2a+9d = 10...(1)
Sum of next ten terms = 250, i.e. sum of first 20 terms = 50+250 = 300
(20/2) (2a+19d) = 300
2a+19d = 30 ...(2)
(2)-(1): 10d = 20
d = 2
a = [10-9(2)]/2 = -4
The 13th term: -4 + (13-1)(2) = -4+24 = 20
2. First term = sum to the 1st term = (3-1)/2 = 1
Sum to the 2nd term = (3^2 - 1)/2 = 4
Hence the second term = 4 - 1 = 3
Therefore, common ratio = 3/1 = 3
General term = 1(3^(n-1)) = 3^(n-1)

1. Given S(10)=50
hence [2a+(10－1)d] x 10 ÷ 2=50
2a+9d=10
2a=10－9d
And S(20)=50+250=300
hence [2a+(20－1)d] x 20 ÷ 2=300
2a+19d=30
10－9d+19d=30
10d=20
d=2
Therefore a= -4
So T(13)= -4+(13－1) x 2=20

2. ∵T(1)=S(1)=1
∴n=1,S(2)－1,S(3)－S(2),S(4)－S(3)...
n=1,3,9,27,...
T(n)=3^(n－1)