Question:
Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.
Solution:
Here, let us take the first term of the A.P as a and the common difference of the A.P as d
Now, as we know,
$a_{n}=a+(n-1) d$
So, for $3^{\text {rd }}$ term $(n=3)$,
$a_{3}=a+(3-1) d$
$16=a+2 d$
$a=16-2 d$.......(1)
Also, for 5th term (n = 5),
$a_{5}=a+(5-1) d$
$=a+4 d$
For 7th term (n = 7),
$a_{7}=a+(7-1) d$
$=a+6 d$
Now, we are given,
$a_{7}=12+a_{5}$
$a+6 d=12+a+4 d$
$6 d-4 d=12$
$2 d=12$
$d=6$
Substituting the value of d in (1), we get,
$a=16-2(6)$
$=16-12$
$=4$
So, the first term is 4 and the common difference is 6.
Therefore, the A.P. is