16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
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Hey User !!!
Let a be the First term, a3 be the third term, a5 be the 5th term and a7 be the 7th term
a3 = 16
a7 = a5 + 12 ............ (1)
Let the common difference be "d"
Common difference is equal in AP
So,
a7 = a5 + d + d = a5 + 2d ............(2)
From Equation (1) & (2)
a5 + 12 = a5 + 2d
2d = 12
d = 6
From Given, we get that
a3 = 16
a3 = a + 2d = 16
a + ( 2 × 6 ) = 16 [ We know that d = 6 ]
a + 12 = 16
a = 4
So first term is 4 .... We can find AP by adding d continuously
So, AP is 4, 10, 16, 22, 28.......
Hope it helps !!!
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