(i) a_{n} = 3 + 4n

(ii) a_{n} = 9 – 5n

Also, find the sum of the first 15 terms in each case.

**Solution:**

**(i) a _{n} = 3 + 4n**

a_{1} = 3 + 4(1) = 7

a_{2} = 3 + 4(2) = 3 + 8 = 11

a_{3} = 3 + 4(3) = 3 + 12 = 15

a_{4} = 3 + 4(4) = 3 + 16 = 19

We can see here, the common difference between the terms are;

a_{2} − a_{1} = 11 − 7 = 4

a_{3} − a_{2} = 15 − 11 = 4

a_{4} − a_{3} = 19 − 15 = 4

Hence, a_{k} _{ + 1} − a_{k} is the same value every time. Therefore, this is an AP with a common difference of 4 and the first term as 7.

Now, we know, the sum of n terms is;

S_{n} = n /2[2a + (n – 1) d ]

S_{15 } = 15/2[2(7) + (15 – 1) × 4]

= 15/2[(14) + 56]

= 15/2(70)

= 15 × 35

= 525

**(ii) a _{n} = 9 − 5n**

a_{1} = 9 − 5 × 1 = 9 − 5 = 4

a_{2} = 9 − 5 × 2 = 9 − 10 = −1

a_{3} = 9 − 5 × 3 = 9 − 15 = −6

a_{4} = 9 − 5 × 4 = 9 − 20 = −11

We can see here, the common difference between the terms are;

a_{2} − a_{1} = −1 − 4 = −5

a_{3} − a_{2} = −6 − (−1) = −5

a_{4} − a_{3} = −11 − (−6) = −5

Hence, a_{k} _{ + 1} − a_{k} is the same every time. Therefore, this is an AP with a common difference of −5 and the first term as 4.

Now, we know, the sum of n terms is;

S_{n} = n /2 [2a + (n – 1) d ]

S_{15 } = 15/2[2(4) + (15 – 1)(-5)]

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

= 15/2(8- 7 0)

= 15/2(-62)

= 15(-31)

= -465