the nth term of ap is an=4n-5 then find thesum of first 25 term
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tn = 4n-5
t1 =a = 4(1)-5 =-1
t2 = 4(2)-5=3
difference (d)= t2-t1= 3-(-1)=4
Sn= n/2 {2a+(n-1)d}
S25= 25/2 {2(-1)+(25-1)4} = 1100
t1 =a = 4(1)-5 =-1
t2 = 4(2)-5=3
difference (d)= t2-t1= 3-(-1)=4
Sn= n/2 {2a+(n-1)d}
S25= 25/2 {2(-1)+(25-1)4} = 1100
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