In an AP the sum of first n terms is 3n2÷2+5n÷2 then find the 25th term.
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 , n th term of A.P. is given as,
an=a+(n−1)d
So ,
a25=4+(25−1)(3)
⇒a25=4+24×3=4+72
Hence, 25th term of the AP is 76