(i) a_n = 3+4n
(ii) a_n = 9−5n
Also find the sum of the first 15 terms in each case.

Solutions:

(i) a_n = 3+4n

a₁ = 3+4(1) = 7

a₂ = 3+4(2) = 3+8 = 11

a₃ = 3+4(3) = 3+12 = 15

a₄ = 3+4(4) = 3+16 = 19

The common difference between the terms may be seen here;

a₂ − a₁ = 11−7 = 4

a₃ − a₂ = 15−11 = 4

a₄ − a₃ = 19−15 = 4

As a result, a_{k + 1} − a_k value is always the same. As a result, this AP has the common difference as 4 and the first term as 7.

We now know that the sum of the nth term is;

S_n = n/2[2a+(n -1)d]

S₁₅ = 15/2[2(7)+(15-1)×4]

= 15/2[(14)+56]

= 15/2(70)

= 15×35

= 525

(ii) a_n = 9−5n

a₁ = 9−5×1 = 9−5 = 4

a₂ = 9−5×2 = 9−10 = −1

a₃ = 9−5×3 = 9−15 = −6

a₄ = 9−5×4 = 9−20 = −11

The common difference between the terms may be seen here;

a₂ − a₁ = −1−4 = −5

a₃ − a₂ = −6−(−1) = −5

a₄ − a₃ = −11−(−6) = −5

As a result, a_{k + 1} − a_k  value is always the same. As a result, this is A.P. has the common difference as −5 and the first term as 4.

We now know that the sum of the nth term is;

S_n = n/2 [2a +(n-1)d]

S₁₅ = 15/2[2(4) +(15 -1)(-5)]

= 15/2[8 +14(-5)]

= 15/2(8-70)

= 15/2(-62)

= 15(-31)

= -465