Arithmetic sequence question HELP

The sum of first nine terms of an arithmetic progression is 75 and the 25th term is also 75.Find the common difference and the sum of the first hundred terms.

(9 / 2)(2a + 8d) = 75
=> a + 4d = 25 / 3
a + 24d = 75
=> 20d = 200/3
=> d = 10/3
=> a = -5
S100 = 50*(-10 + 99 * 10 / 3)
=> S = 50*(320)
=> S = 16000

Arithmetic sequence is like, starting with a number a, and add a certain value

a , a+r , a+2r , a+3r , ...
the general term

Tn = a + (n-1) r , a is the first term and r is the value we add and n is the number of the term

sum of first 9 terms

a + (a + r) + (a + 2r) + ... + (a + 8r) = 9a + ( r + 2r + 3r + ... + 8r) = 9a + r(1 + 2 + ... +8) = 9a + 36r

9a+36r = 75 --(1)

they said the 25th term is 75

T25 = a + (25 - 1 )r = a+24r = 75 --(2)
a = 75 - 24r.

9(75 - 24r) + 36 r = 75
9(75) - 9(24)r + 36r = 75
36 r - 9(24)r = 75 - 9(75)
9(4r - 24r) = -8(75)
9(-20r) = -8(75)
r = 8(5.15)/(4.5.9) = (2.5)/(3) = 10/3

a = 75 - 24(10/3) = 75 - 8(10) = 75 - 80 = -5
General term
Tn = -5 + (n -1 )(10/3)

first 100 term sum
the sum of the nth term

Sn = na + (n-1)n(r)/2

S100 = 100(-5) + (100)(99)(10/3)/2

S100 = 100 ( -5 + 33(5)) , taking 100 as a factor canceling 3 with 99 and 2 with 10

S100 = 100 (5( -1 + 33)) = 100(5(32)) = 500(32) = 1600