The su吗the sum of the 1st twenty-one of linear sequence is 28, and sum of the first twenty -eight terms is 21.find (a) the term of the sequence that is equal to 0. (b) the sum of the terms preceding the term 0

Answer 1

sum(21) = 28
(21/2)(2a + 20d) = 28
2a+20d = 56/21

sum(28) = 21
(28/2)(2a + 27d) = 21
2a+27d = 42/28 = 3/2

subtract the two equations:
7d = 3/2 - 56/21 = -7/6
d = -1/6

in 2a + 20d = 56/21
2a - 20/6 = 56/21
2a = 6
a = 3

so we need:
a + (n-1)d = 0
3 + (n-1)(-1/6) = 0
3 - (-1/6)n + 1/6 = 0
19/6 = n/6
n = 19

sum(19) = 0

check:
a+18d
= 3 + 18(-1/6) = 0

b)
so sum(18) = ......... (your turn)

Answer 2

S18 =18÷2(2a+[n-1]d).
S18=9(6+[17]-1÷6
S18=9(6-17÷6)
S18=28.53