Question

Find the sum of first 25 terms of an A.P. whose 'n'th term is 1-4n.

Answer 1

put n=1 2 &3
then series become -3 -7 -11 as so
here a=-3
d=-4
n=25
S25=n/2[2a+(n-1)d]
25/2[2×(-3)+(25-1)-4]
25×(-51)= -1275

Answer 2

Give some value for n.
If n,=1. 1-4(1)= 1-4=3
If n=2. 1-4(2)=1-8=7
if n=3. 1-4(3)=1-12=11

Here, AP= 3,7,11,.......
D= 7-3= 4

So here, a=3,d=4,n=25
put values in,
Sn=n/2(2a+(n-1)d)
S25 = 25/2(2*3+(25-1)4)
= 25/2(6+24*4)=25/2(6+96)
. =25/2(102)=25*51=1275 ans

Hope it help........