Question

In an AP , the sum of first terms is ( 3n²/2 + 5n/2 ) . find the 25th term..

Answer 1

It is given that Sn = ( 3N² /2 + 5n /2 ) --------(1)


Now , 25th term = ( Sum of first 25 terms ) - ( sum of first 24 terms ).


Putting n = 25 in (1) , we get



S25 = ( 3 × (25)² /2 + 5 × 25 /2 )


S25 = ( 3 × 625 /2 + 125/2 )


S25 = ( 1875 /2 + 125/2 )


S25 = 2000/2 = 1000.


Putting N = 24 in (1) , we get


S24 = ( 3 × (24)² /2 + 5 × 24 /2 )


S24 = ( 1728/2 + 120/2 )


S24 = (1848/2 ) = 924.


Therefore,


T25 = ( S25 - S24 )


T25 = ( 1000 - 924 )


T25 = 76.


Hence,


The 25th term is 76.

Answer 2

Let the sum of n terms be given by Sn
so
Sn = 3n²/2+ 5n/2
S1 = 3(1)²/2 + 5(1)/2
= 3/2+5/2 => 4
so 1st term is 4 say 'a'
Now
S2 = 3(2)²/2 + 5(2)/2
= 6+5 => 11
Now a2 = S2 - a1
=> a2 = 11-4 = 7
Now common difference (d)
= a2-a1 => 7-4 = 3
we know
an = a+(n-1) d
so
a25 = 4 + (25-1)(3)
=> a25 = 4 + 24×3 =>4+ 72
so 25th term of the AP is 76.