Sum

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

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#### Solution

Given:

3rd term of the AP is 16.

a_{3} = 16

a + (3 − 1)d = 16

a + 2d = 16 .....(1)

Also, 7th term exceeds the 5th term by 12.

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, …

Concept: nth Term of an AP

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