# the pth term of A.P. is (3p-1)/6.The sum of the first n terms of the AP ?

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• rotchm
Lv 7
1 month ago

Did you see a formula for the sum? If so, show it here & apply it if u can.

(Note that they say that the first term is p=1. The starting term is important & needs to be specified, which it is here.)

Or, its

n

∑ (3p-1)/6 =

p=1

(1/6)( 3∑p - ∑1 ). Convince yourself of this.

∑1 = 1+1+1+...+1 , n times. Thus = n.

∑p = 1+2+3+... + n = n*(n+1)/2.  Why? [you should know this one by heart. If not, google/wiki triangular numbers).

What do you finally get?

• 1 month ago

The first term (p = 1) is a = (3*1 - 1) / 6 = (3 - 1) / 6 = 2/6 = 1/3

The common difference is term p+1 mins term p:

d = ((3(p + 1) - 1) / 6) - ((3p - 1) / 6)

d = ((3p + 3 - 1) / 6) - ((3p - 1) / 6)

d = ((3p + 2) / 6) - ((3p - 1) / 6)

d = ((3p + 2) - (3p - 1)) / 6

d = (3p + 2 - 3p + 1) / 6

d = 3/6

d = 1/2

The sum of the first n terms is:

S = (n/2)(2a + (n - 1)*d)

S = (n/2)(2(1/3) + (n - 1)(1/2))

S = (n/2)((2/3) + (1/2)n - (1/2))

S = (n/2)((1/2)n + (1/6))

S = (1/4)n^2 + (1/12)n

S = (3n^2 + n) / 12

• 1 month ago

With p = 1, we have 2/6 = 1/3

With p = 2, we have 5/6

So, the arithmetic series has 1st term 1/3 and common difference 1/2

Now, Sₙ = (n/2)[2a + (n - 1)d]

Hence, Sₙ = (n/2)[2/3 + (1/2)(n - 1)]

so, Sₙ = (n/2)[1/6 + n/2] = (n/12)(1 + 3n)

:)>