Answer: (a) -14 + 15n - 15, (b) 670
Explanation:
The nth term of an Arithmetic sequence is
Tn = a + ( n - 1 )d
From the question, the 12th term = 41 and the 4th term = 1, to find the a the first term and d the common difference of the sequence, requires a little understanding. Now, we have to resolve to a simultaneous equation to get the two unknown. Now let's go:
From.the first statement
T12 = a + ( n - 1 ) = 41 --------------- 1,
Second statement,
T4 = a + ( n - 1 ) = 1 ------------------ 2
Now solve the two equations together using any known methods
T12 = a + 11d = 41
T4 = a + 3d = 1
-------------- , now subtract.
8d = 40 and
d = 5.
To find the value of a, in oder to proceed to the answers to the question, substitute for d in any of the equation above.
a + 3d = 1
a + 3 × 5 = 1
a + 15 = 1
a = 1 - 15
a = -14
So a = -14 and d = 5, Now answers to the questiona are
(a) Tn = a + ( n - 1 )d
= -14 + ( n - 1 )5
= -14 + 15n - 15
(b) sum of the 20 terms.
S20 = n/2{(2a + ( n - 1 )d )}
= 20/2{( 2× (-14) + (20 -1) × 5
= 10(-28 + 19 × 5 )
= 10( -28 + 95)
= 10( 67)
= 670.