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Answer:

A sequence that has common difference between any two of its consecutive terms is an arithmetic progression.

Sum of the first n terms of an AP is given by Sₙ = n/2 [2a + (n - 1) d] or Sₙ = n/2 [a + l], and the nth term of an AP is aₙ = a + (n - 1)d

Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.

(i) aₙ = 3 + 4n

Given,

nth term, 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

It can be observed that

a₂ - a₁ =11 - 7 = 4

a₃ - a₂ =15 - 11 = 4

a₄ - a₃ =19 - 15 = 4

So, the difference of aₙ and aₙ₋₁ is constant.

Therefore, this is an AP with common difference as 4 and first term as 7.

The sum of n terms of AP is given by thre formula Sₙ = 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) aₙ = 9 - 5n

Given,

nth term is 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

It can be observed that

a₂ - a₁ = (-1 ) - 4 = - 5

a₃ - a₂ = (-6) -(-1) = - 5

a₄ - a₃ = (- 11) - (- 6) = - 5

So, the difference of aₙ and aₙ₋₁ is constant.

Therefore, this is an A.P. with common difference - 5 and first term as 4.

The sum of n terms of AP is given by thre formula Sₙ = 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)

= - 465

Explanation:

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