find the AP whose third term is 16 and 7th term is 40
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Heya !!
Here is your answer..
Third term = 16 = a + 2d -----(i)
Seventh term = 40 = a + 6d -----(ii)
(where d is difference and a is first term)
Subtracting (i) from (ii), we get
==> 4d = 24
==> d = 6
Also, a + 2d = 16
==> a = 16 - 12
==> 4
Now, the required AP is
4, 10, 16, 22
Hope it helps..
Here is your answer..
Third term = 16 = a + 2d -----(i)
Seventh term = 40 = a + 6d -----(ii)
(where d is difference and a is first term)
Subtracting (i) from (ii), we get
==> 4d = 24
==> d = 6
Also, a + 2d = 16
==> a = 16 - 12
==> 4
Now, the required AP is
4, 10, 16, 22
Hope it helps..
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