Question

solve this....
NO SPAMMING...
​

Answer 1

Concept:-Solve for n=1 to n=infinity.....All the other values of n will lie between them

Solution:-clearly from the given pattern we can say that

a1=√7......i)

a1=2.4(approx)

now if n=infinite,,then

$a \infty = \sqrt{7 + \sqrt{7 + \sqrt{7 + ...... + \infty } } } \\ {(a \infty )}^{2} = 7 + \sqrt{7 + \sqrt{7 + ..... + \infty } } \\ subtracting \: we \: get \\ {(a \infty )}^{2} - a \infty = 7 \\ a \infty = \frac{ 1 + - ( \sqrt{ {1}^{2} + 28 } }{2} \\ a \infty = \frac{1 + \sqrt{29} }{2} = \frac{6.3}{2} = 3.15$

and,as mentioned above

$2.4 \leqslant an \leqslant 3.15$

option C is saying that an can never be greater than 4,,,which is true

other options are not following the condition ,which we get ,so

Option c)✔️✔️