8th term is 31 , 5th term is 16 more than 7th term find AP
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since A.P series is like this
a, (a+d), (a+2d), (a+3d), (a+4d), ……………………………. , {a+(n-1)d}
and nth term is given by = {a+(n-1)d}
so 8th term will be a+7d ,
11th term will be a+10d and
15th term will be a+14d
so as per qn.
a+7d= 31 ……………… (1)
& a+14d = a+10d +16
therefore 4d = 16
& d = 4 ……………….. (2)
use by using (2) in (1)
a + 7 *4 = 31
we get a = 31–28 = 3
so the required A.P will be 3, 7, 11, 15, 19, 23,
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