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What will be the sum of ten terms of an Arithmetic progression whose 5th term is 5 and 7th term is 3?
Wiki User
∙ 10y ago
U5 = a + 4d = 5
U7 = a + 6d = 3
Subtracting the first equation from the second gives
2d = -2 therefore d = -1
and then a + 4d = 5 implies that a = 9.
So Sn = n/2*{2a + (n-1)*d}
thus S10 = 10/2*{18 + 9*(-1)}
= 5*{9} = 45
Wiki User
∙ 10y ago
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