Show that a₁ , a₂ , . . ., a_n , . . . form an AP where a_n is defined as below:

(i) a_n = 3 + 4n (ii) a_n = 9 – 5n 

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

(i) a_n = 3 + 4n

Given, a_n = 3 + 4n

Here,

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

Also the common difference,

d= a₂ - a₁ =11 - 7 = 4

d= a₃ - a₂ =15 - 11 = 4

d= a₄ - a₃ =19 - 15 = 4

So, the difference is constant. 

Therefore, this is an AP with d = 4 and a = 7.

Using the formula,

S_n = n/2 [2a + (n - 1)d]

Sum of 15 terms,

S₁₅ = 15/2 [2 × 7 + (15 - 1)4]

= 15/2 [14 + 14 × 4]

= 15/2 × 70

= 15 × 35

= 525

(ii) a_n = 9 - 5n

Given,  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

Now we get,

d = a₂ - a₁ = (-1) - 4 = - 5

d = a₃ -a₂ = (-6) -(-1) = - 5

d = a₄ - a₃ = (- 11) - (- 6) = - 5

So, the difference is constant.

Therefore, this is an A.P. with d = - 5 and a = 4.

Using the formula,

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)

S₁₅ = - 465