You are right the answer is A
Answer:
First term a = 3
Sum of first 20 term = 440
Step-by-step explanation:
Given:
8th term of AP = 17
12th term of AP = 25
Find:
First term a
Sum of first 20 term
Computation:
8th term of AP = 17
a + 7d = 17 ....... EQ1
12th term of AP = 25
a + 11d = 25 ...... EQ2
From EQ1 and EQ2
4d = 8
d = 2
a + 7d = 17
a + 7(2) = 17
First term a = 3
Sum of first 20 term
Sn = [n/2][2a + (n-1)d]
S20 = [20/2][(2)(3) + (20-1)2]
S20 = [10][(6) + 38]
S20 = [10][44]
S20 = 440
Sum of first 20 term = 440
Answer:
Part A: b < 28
Part D: n < 56
Step-by-step explanation:
part A: divide each term in and simplify.
Inequality Form: b < 28
part d: this is already solved but can be shown in multiple forms.
Inequality Form: n < 56
Interval Notation: (
-∞
, 56)
