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Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.
The 3rd term of an A.P. is 16 and its 7th term exceeds the 5th term by 12.
We have to find the AP.
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. $a_n=a+(n-1)d$
According to the question,
$a=16-2(6)$ (From (i))
$a_1=a=4, a_2=a+d=4+6=10, a_3=a+2d=4+2(6)=4+12=16$
Hence, the required arithmetic progression is $4, 10, 16, .......$
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