the sum of the first n terms of an AP is given by sn=3n2-n , determine the AP and its 25thterm.

Dear Student!

Given, sum of first n terms of an AP, Sn = 3n2n

 Sn = 3n2 n

Replacing n by n – 1, we get

Let the nth term of AP be an.



Now,put n = 1 in an, we get a = 6×1-4 = 2put n = 2 in an, we get a2 = 6×2-4 = 8put n = 3 in an, we get a3 = 6×3-4 = 14and so on.So, required AP is, 2,8,14,.....

Putting n = 25, we get

a25 = 6 × 25 – 4 = 150 – 4 = 146

Thus, the 25th term of the given A.P. is 146.

Cheers!

  • 52

given sum of first n terms of a given terms:

Sn =3n2 - n

giving some values for n we get:

if n=1

s1 = 3*12 - 1

  = 3-1

  = 2

so we get the sum of 1 term = 2.in fact the first term = 2.

if n=2:

s2 = 3*22 - 2

  = 3*4 - 2

  = 12-2

  = 10

so the sum of first two terms = 10.

so the second term = sum of two terms - the first term

  = 10 - 2

  = 8.

so d = 8-2

  = 6

so the required AP is:

2,2+6,2+(2*6),2+(3*6)...

2,8,14,20,26,32...

a25 = a+19d

  = 2 + 19*6

      = 2 + 114

a25 = 116 

  • 26
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