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) Given,

⇒ a_n = 3 + 4n

⇒ a₁ = 3 + 4 × 1 = 3 + 4 = 7

⇒ a₂ = 3 + 4 × 2 = 3 + 8 = 11

⇒ a₃ = 3 + 4 × 3 = 3 + 12 = 15

a₂ - a₁ = 11 - 7 = 4,

a₃ - a₂ = 15 - 11 = 4.

Since, a₃ - a₂ = a₂ - a₁.

Hence, it is an A.P. with common difference = 4 and first term (a₁) = 7.

a₁₅ = 3 + 4 × 15 [∵ a_n = 3 + 4n]

= 3 + 60 = 63.

By formula,

Sum of A.P. = $\dfrac{n}{2}$ [First term + Last term]

Substituting values we get :

$S_{15} = \dfrac{15}{2}[7 + 63] \\[1em] = \dfrac{15}{2} \times 70 \\[1em] = 15 \times 35 \\[1em] = 525.$

Hence, sum of first 15 terms = 525.

(ii) 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

a₂ - a₁ = -1 - 4 = -5,

a₃ - a₂ = -6 - (-1) = -5.

Since, a₃ - a₂ = a₂ - a₁.

Hence, it is an A.P. with common difference = -5 and first term (a₁) = 4.

a₁₅ = 9 - (5 x 15) [∵ a_n = 9 - 5n]

= 9 - 75

= -66.

By formula,

Sum of A.P. = $\dfrac{n}{2}$ [First term + Last term]

Substituting values we get :

$S_{15} = \dfrac{15}{2}[4 + (-66)] \\[1em] = \dfrac{15}{2} \times -62 \\[1em] = 15 \times -31 \\[1em] = -465.$

Hence, sum of first 15 terms = -465.