Arun
Last Activity: 7 Years ago
S= (n/2)[2a+ (n–1) d]
S= na + (n2–n)*d/2
S= (a –d/2)n + n2d/2
now we compare the coefficient of n2 & n.
d=6, a –3= 2
d =6, a =5
hence a and d for second AP
a= 5, d = 12
Sn= (n/2)[10+(n –1)*12]
Sn = 5n + 6n2 –6n
Sn= 6n2 –n
(
answer)
S= (a –d/2)n + n2d/2
since we know that for second AP d is twice and a is same
hence put for S for secnd AP, d= 2d
Sn = (a –d)n + n2d
put the value of a= 5, d = 6
S
n = 6n
2 –n
Another method
S1 = a = T1 = 2+3= 5
now if we find S2 = T1+T2 = 16
hence T2= 16 –5 = 11
a+d = 11
d= 11–5 = 6
now we have find a &d for the first AP.
for second AP,
a = 5, d= 12
so Sn for second AP = (n/2)[2*5 + (n –1)* 12]
Sn = 5n + 6n2 –6n
Sn = 6n2 –n
