(i) an = 3 + 4n
(ii) an = 9 – 5n
Also, find the sum of the first 15 terms in each case.
Solution:
(i) an = 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
We can see here, the common difference between the terms are;
a2 − a1 = 11 − 7 = 4
a3 − a2 = 15 − 11 = 4
a4 − a3 = 19 − 15 = 4
Hence, ak + 1 − ak 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;
Sn = 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) an = 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
We can see here, the common difference between the terms are;
a2 − a1 = −1 − 4 = −5
a3 − a2 = −6 − (−1) = −5
a4 − a3 = −11 − (−6) = −5
Hence, ak + 1 − ak 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;
Sn = n /2 [2a + (n – 1) d ]
S15 = 15/2[2(4) + (15 – 1)(-5)]
= 15/2[8 + 14(-5)]
= 15/2(8- 7 0)
= 15/2(-62)
= 15(-31)
= -465