The $$3^{rd}, 14^{th}$$ and $$69^{th}$$ terms of an arithmetic progression form three distinct and consecutive terms of a geometric progression. If the next terms of the geometries progression is the $$n^{th}$$ term of the arithmetic progression, then n quals____________.

Let the first term of arithmetic progression be a and common difference be d.

So, $$3^{rd}$$ term = $$a+2d$$, $$14^{th}$$ term = $$a+13d$$, $$69^{th}$$ term = $$a+68d$$.

These terms are in G.P.

So, $$\left(a+13d\right)^2=\left(a+2d\right)\cdot\left(a+68d\right)$$

or, $$a^2+26ad+169d^2=a^2+2ad+68ad+136d^2$$

or, $$169d^2-136d^2=70ad-26ad$$

or, $$33d^2=44ad$$

or, $$3d=4a$$

or, $$a=3d/4$$

So, $$3^{rd}$$ term =$$a+2d=\dfrac{3d}{4}+2d=\dfrac{11d}{4}$$

$$14^{th}$$ term =$$a+13d=\dfrac{3d}{4}+13d=\dfrac{55d}{4}$$

$$69^{th}$$ term =$$a+68d=\dfrac{3d}{4}+68d=\dfrac{275d}{4}$$

We can see that the common ratio of this GP is 5

So, next term of this G.P. will be $$\dfrac{275d}{4}\cdot5=\dfrac{1375d}{4}$$

And, $$\dfrac{1375d}{4}=a+\left(n-1\right)d=\dfrac{3d}{4}+\left(n-1\right)d$$

or, $$\dfrac{1375d}{4}-\dfrac{3d}{4}=\left(n-1\right)d$$

or, $$\dfrac{1372d}{4}=\left(n-1\right)d$$

or, $$n-1=343$$

or, $$n=344$$