#### Question

Determine the A.P. whose third term is 16 and the 7^{th} term exceeds the 5^{th} term by 12.

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#### Solution

=a_{3} = 16

a + (3 − 1) d = 16

a + 2d = 16 (1)

a_{7} − a_{5} = 12

[a+ (7 − 1) d] − [a + (5 − 1) d]= 12

(a + 6d) − (a + 4d) = 12

2d = 12

d = 6

From equation (1), we obtain

a + 2 (6) = 16

a + 12 = 16

a = 4

Therefore, A.P. will be

4, 10, 16, 22, …

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#### Reference Material

Solution for question: Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12. concept: null - nth Term of an AP. For the course CBSE