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

Given that

Third term, a3 = 16

The 7th term exceeds the 5th term by 12=>a7 − a5 = 12

Find out

We have to find out the AP series from the given data

Solution

We know that

a +(3−1)d = 16

a+2d = 16 ………………………………………. (i)

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

a7 − a5 = 12

[a+(7−1)d]−[a +(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

a2= 4+ 6 = 10 

a3= 10+ 6 = 16

Therefore, A.P. will be 4, 10, 16, 22, …

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