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Q.
If the greatest divisor of $30 \cdot 5^{2 n}+4 \cdot 2^{3 n}$ is $p, \forall n \in N$ and that of $2^{2 n+1}-6 n-2$ is $q, \forall n \in N$, then $p+q=$
TS EAMCET 2019
Solution:
We have, $30 \cdot 5^{2 n}+4 \cdot 2^{3 n}$
Put $n=1$
$30 \times 25+4 \times 8=750+32=782$
$782=2 \times 23 \times 17\,\,\,\,\,\dots(i)$
Put $x=2$
$30 \times 625+4 \times 64=18750+256=19006\,\,\,\,\,\dots(ii)$
In Eqs. (i) and (ii), we get
Greatest divisor $=2 \times 17=34$
$\therefore \, p=34$
Now in $2^{2 n+1}-6 n-2$
Put $n=1$
$8-6-2=0$
Put $n=2 $
$ 32-12-2=18 $
Put $n=3$
$ 128-18-2= 108 $
$ \because$ Greatest divisor $= 18 $
$ \therefore \, q=18 $
$\therefore \,p+q=34+18=52$