what is the 3rd termof an Arithmetic progression whose 4th term is 9 and common diffrence is 2
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Answer:T4 = a+3d = 13 …(1)
T10 = a+9d = 25 …(2)
Subtract (1) from (2)
a+9d-a-3d = 6d = 25–13 = 12
So d = 2.
From (1), a = 13 - 6 = 7
The first term is 7, the common difference is 2 and the 17th term = a+16d = 7+32 = 39.
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