Question

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​

Answer 1

To show that the terms of the series with sum Sn = n(3n-4) are in arithmetic progression, we find the nth and (n+1)th terms and demonstrate that their difference is constant. This confirms the series as an arithmetic progression.

Step-by-step explanation:

The student is asking to show that the terms of the series with the sum Sn = n(3n-4) are in arithmetic progression (AP). To verify this, we consider the difference between sums of consecutive terms. The sum of the first n terms is given by Sn, so the nth term of the series, Tn, can be calculated as Tn = Sn - Sn-1.

Let's calculate Tn and Tn+1 step by step:

• The sum of the first n terms: Sn = n(3n-4)
• The sum of the first (n-1) terms: Sn-1 = (n-1)(3(n-1)-4)
• The nth term: Tn = Sn - Sn-1
• The (n+1)th term: Tn+1 = Sn+1 - Sn

Now, we calculate Tn and Tn+1 and find their difference to determine if it is constant:

• Tn = n(3n-4) - (n-1)(3(n-1)-4)
• Tn+1 = (n+1)(3(n+1)-4) - n(3n-4)

After simplifying both Tn and Tn+1, we see that the difference Tn+1 - Tn is constant (6), indicating that the terms of the series are indeed in an arithmetic progression.

Answer 2

see explanation

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