Answer:
Explanation:
1) we want to find the 12th term of the arithmetic sequence given that the 7th term is 40 and the 18th term is 106
Recall
Tn = a + (n-1)d
Where Tn is the value of the nth term
a is the value of the first term
d is the common difference between a term and the following term
n is the number of terms
For the 7th term,
40 = a + (7-1)d
40 = a + 6d - - - - - - -1
For the 18th term,
106 = a + (18-1)d
106 = a + 17d - - - - - - -2
Subtract equation 2 from equation 1
-66 = -11d
d = 66/11 = 6
Put d = 6 in equation 1
a = 40-6d
a = 40 - 6×6
a = 40 - 36 = 4
To find the 7th term,
T 12 = 4 + (12-1)6
T12 = 4 + 11×6 =4 +66 =70
2) we want to find the 7th term of the geometric sequence given that the 2nd term is 324 and the 4th term is 36. Recall
Tn = ar^(n-1) where
a is the value of the first term
r is the common ratio between a term and the previous term
n is the number of terms.
For 2nd term
324 = ar^(2-1)
324 = ar- - - - - - -1
For 4th term
36 = ar^(4-1)
36 = ar^3 - - - - - -2
Divide equation 1 by equation 2
324/36 = ar / ar^3
9 = r^(1-3) = r^-2
9 = 1/r^2
9r^2 = 1
r^2 = 1/9
Take square root of both sides
r = 1/3
Put r = 1/3 in equation 1
324 = 1/3 × a
a/3 = 324
a = 3 × 324 = 972
To find the 7th term,
T7 = 972 × 1/3^(7-1)
T7 = 972 × (1/3)^6
T7 = 972 × 0.00137174211
T7 = 4/3
7th term is 4/3