Question

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
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Answer 1

Answer:

3rd term of AP = 16

⇒ a + (n-1)d = 16

a + (3-1)d = 16

a + 2d = 16

a = 16 - 2d — (i)

5th term of AP = y

⇒ a + (n-1)d = y

a + (5-1)d = y

a + 4d = y

7th term of AP = y + 12

⇒ a + (n-1)d = 12 + y

a + (7-1)d = 12 + y

a + 6d = 12 + a + 4d [y = a + 4d]

⇒ a + 6d = 12 + a + 4d

a + 6d - a - 4d = 12

2d = 12

d = 6

Now we find the value of ‘a’ by substituting the value of ‘d’ in equation (i):-

a = 16 - 2d

a = 16 - 2(6)

a = 16 - 12

a = 4

AP formed:-

4, 10, 16, 22, 28, 34, 40..............

Hope it helps

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Answer 2

Answer:

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