Consider the arithmetic sequence where the 12th term is 41 and the 4th term is 1.
a. Find the formula of the nth term of the sequence.

b. Find the sum of the first 20 terms.

(a) -14 + 15n - 15, (b) 670

Step-by-step 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.