Sum of first 12 term of an A.P. Is 28 & Sum of first 28 term Is 12 find sum of 40th term.
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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
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
Wrong
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