Question
Sun March 14, 2010 By: Lausonia R. Swamy

pls answer quickly tuesday is exam

Expert Reply
Mon March 15, 2010

sn = n(2a0 + (n-1)d)/2            ... first AP

and

Sn =n(2A0 + (n-1)D)/2         ... second AP.

sn/Sn = (2a0 + (n-1)d)/(2A0 + (n-1)D) = (3n+8)/(7n+15)

Put n = 23,

s23/S23 = (2a0 + 22)d)/(2A0 + 22D) = (3x23+8)/(7x23+15)  = 77/176

(a0 + 11)d)/(A0 + 11D) = (3x23+8)/(7x23+15)  = 77/176

a12/A12 =  (a0 + 11d)/(A0 + 11D) = 77/176

Regards,

Team,

TopperLearning.

 

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