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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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