JEE Main 2023 — Mathematics Algebra

Given that $n\in [10,100]$n∈[10,100] and $n\in N$n∈N.

Let us take the values of $n=1,2,3,...$n=1,2,3,... and check whether ${3}^{n}-3$3n−3 is divisible by $7$7 or not.

$\Rightarrow {3}^{1}-3=0$⇒31−3=0 is divisible by $7$7

$\Rightarrow {3}^{2}-3=3$⇒32−3=3, $3$3 is not divisible by $7$7

$\Rightarrow {3}^{6}=7k+1$⇒36=7k+1 is not divisible by $7$7

$\Rightarrow {3}^{7}=7\alpha +3$⇒37=7α+3,

$\Rightarrow {3}^{7}-3=7\alpha$⇒37−3=7α is divisible by $7$7

Similarly $\Rightarrow {3}^{13}=(7k+1)(7\alpha +3)$⇒313=(7k+1)(7α+3)

$$=7\beta +3$$=7β+3

$\Rightarrow {3}^{13}-3=7\beta$⇒313−3=7β is divisible by $7$7

If we observe the pattern for $n=1,7,13$n=1,7,13, ${3}^{n}-3$3n−3 is divisible by $7$7.

That means the series forms an AP with $d=6$d=6.

Also $n\in [10,100]$n∈[10,100]

The required progression is $13,19,......{a}_{n}$13,19,......an​

$$\Rightarrow {a}_{n}=13+(n-1)6$$⇒an​=13+(n−1)6

But $13+(n-1)6<100$13+(n−1)6<100

$$\Rightarrow n<15.5$$⇒n<15.5

$$\Rightarrow n=15\in N$$⇒n=15∈N

Therefore, the required answer is $15$15.