sum of the first nth term of a series is given by Sn = n (3n-4). Show that the terms of the series are in arithmetic progression​

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Step-by-step explanation:

Using the sum to n terms formula, then

S₁ = 1(3 - 4) = 1(- 1) = - 1 ⇒ a₁ = - 1

S₂ = 2(6 - 4) = 2(2) = 4, then

a₂ = S₂ - S₁ = 4 - (- 1) = 4 + 1 = 5

S₃ = 3(9 - 4) = 3(5) = 15, then

a₃ = S₃ - S₂ = 15 - 4 = 11

S₄ = 4(12 - 4) = 4(8) = 32 , then

a₄ = S₄ - S₃ = 32 - 15 = 17

S₅ = 5(15 - 4) = 5(11) = 55 , then

a₅ = S₅ - S₄ = 55 - 32 = 23

The first 5 terns of the series are

- 1, 5, 11, 17, 23

There is a common difference (d) between consecutive terms

d = 5 - (- 1) = 11 - 5 = 17 - 11 = 23 - 17 = 6

Thus the series is in arithmetic progression

your answer is no

step-by-step explanation: