Question

In an AP,the sum of first n term is 3n2/2+13n/2 find its 25th term

Answer 1

given
Sn=3n^2/2+13n/2
so,
S (n-1)=3 (n-1)^2/2+13 (n-1)/2
we know
tn=Sn-S(n-1)
=3n^2/2+13n/2-3 (n-1)^2/2-13 (n-1)/2
=3n+5
hence tn=3n+5
t25=3 (25)+5=80

Answer 2

given Sn = 3n²/2 +13n/2

s1 = 3*1/2 +13*1/2 = 3/2 +13/2 = 16/2 = 8

s2 = 3*2²/2 + 13*2/2 =6 +13 =19

s3 = 3*3²/2 +13*3/2 = 27/2 +39/2 =66/2 =33

first term a1 = s1 = 8
a2 = s2-s1 =19-8= 11
a3= s3-s2 =33-19 =14
8,11,14,.. are in Ap
a= 8 , common differece = d= a2-a1 = 11-8 =3
nth term = an = a+(n-1)d
a25 = 8+(25-1)3
=8+24*3
= 8 + 72
=80
25 th term in ap is 80