Question

The sum of first n terms of an AP is 3n square - 5 . Find the third term of the AP.
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Answer 1

for the first term we will put n = 1

3 × 1² - 5 = -2

Sum of three terms = n/2(t1 + t3)= 3n²-5

3/2(-2 + t3) = 3×3²-5

22×2/3 + 2 = t3

t3 = 50/3

PLZ MARK AS BRAINLIEST.

Answer 2

Explanation:

General term of the A.P. is 3n^2 - 5

So

T1 = 3 ( 1 )^2 - 5

= 3 -5

= - 2

T2 = 3 ( 2 )^2 - 5

= 12 - 5

= 7

d = 7 - ( - 2 )

= 9

Sn = n/2 [ 2a + ( n - 1 )d ]

= n/2 [2 ( - 2 ) + ( n - 1 )9 ]

= n/2 [ - 4 + 9n - 9 ]

= n/2 ( 9n - 13 )

So the sum of first n terms of A.P. is n/2( 9n - 13 )

T3 = 3 ( 3 )^2 - 5

= 27 - 5

= 22

So the third term of the A.P. is 22

THANKS!!!!