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

Given:

The 3rd term of an A.P. is 16 and its 7th term exceeds the 5th term by 12.

To do:

We have to find the AP.

Solution:

Let the first term, common difference and the number of terms of the given A.P. be $a, d$ and $n$ respectively.

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

$a_{3}=a+(3-1)d$

$16=a+2d$

$a=16-2d$.....(i)

$a_{7}=a+(7-1)d$

$=a+6d$....(ii)

$a_{5}=a+(5-1)d$

$=a+4d$....(iii)

According to the question,

$a_{7}=a_5+12$

$a+6d=(a+4d)+12$

$a+6d-a-4d=12$

$2d=12$

$d=6$

$a=16-2(6)$ (From (i))

$a=16-12$

$a=4$

Therefore,

$a_1=a=4, a_2=a+d=4+6=10, a_3=a+2d=4+2(6)=4+12=16$

Hence, the required arithmetic progression is $4, 10, 16, .......$

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