answer this please guys
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Let say AP is
a , a + d , a + 2d , a + 3d , a + 4d
Sn = (n/2)(2a + (n-1)d )
Sum of 5 terms
= (5/2)(2a + 4d) = 55
=> a + 2d = 11
fourth term is five more than the sum of the first two terms.
=> a + 3d = 5 + a + a + d
=> 2d - a = 5
Adding both
=> 4d = 16
=> d = 4
=> a = 3
AP
3 , 7 , 11 , 15 , 19
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