Question

sum of first n terms of an a.p. is n÷2(3n+13). find its 30th term.

Answer 1

As we know that
T(n) =S(n)-S(n-1
T(30)= S(30)-S(29)
====> {30(3×30+13)/2} -{29(29×3+13)/2}
===> 1545- 1450
T(30)=95

Answer 2

n/2(3n+13)
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put n = 1

1/2(3+13)

=S1=8
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put n =2

2/2(6 +13)

S2 = 19
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put n=3

3/2(9 +13)

S3 = 33
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first term of the AP is 8 
second term = 19 - 8 = 11
third term = 33 - 19 = 14
----
therefore the AP is  8 , 11 , 14 ............

first term = 8
common diff = 3
n = 30

a30 = a +(n-1)d

= 8 + (30-1)3

= 8 + 29(3)

8 + 87

= 95