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Question -

Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.



Answer -

Given,

Third term, a3 = 16

As we know,

a +(3−1)d =16

a+2d = 16………………………………………. (i)

It is given that, 7th termexceeds the 5th term by 12.

a7 − a5 = 12

[a+(7−1)d]−[+(5−1)d]= 12

(a+6d)−(a+4d) = 12

2d = 12

d = 6

From equation (i), we get,

a+2(6) = 16

a+12 = 16

a = 4

Therefore, A.P. will be4, 10, 16, 22, …

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