Answers

2016-04-07T18:53:03+05:30
SUM OF FIRST 12 TERM =n/2[2a + (n-1)d]
n=12   so  28=12/2[2a + (12-1)d]
6[2a+11d]=28   12a+66d = 28     6a+33d=14

12=28/2[2a+(28-1)d]
12=14[2a+27d]
6=14a+189d

solve 6a+33d=14  &  6=14a+189d 
 from this we get 
a=-5/21 
d=36/77
sum of first 40th term =40/2[2*-5/21+ 39*36/77]

SUM OF first 40th term=355.1515  
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