In an arithmetic progression (AP), the 9th term is 5 times the 2nd term and the 8th termis 1 more than 10 times the first term. What is the 4th term of the geometricprogression (GP) whose first term is the second term of AP and whose common ratiois equal to the common difference of AP?

Correct option is A

Given:
In an arithmetic progression (AP), the 9th term is 5 times the 2nd term.
The 8th term is 1 more than 10 times the first term.​​
Formula Used:
For AP (Arithmetic Progression):
​$$T_n = a + (n-1)d$$​​
For GP (Geometric Progression):
​$$T_n = T_1 \times r^{n-1}$$​​
Solution:
Let the first term of the AP be a and the common difference be d.
From the given, the 9th term is 5 times the 2nd term:
a + 8d = 5(a + d)
a + 8d = 5a + 5d
​$$\Rightarrow \quad 3d = 4a$$​​
​$$\Rightarrow \quad d = \frac{4a}{3}$$​​
The 8th term is 1 more than 10 times the first term:
a + 7d = 10
Substitute $$d = \frac{4a}{3}:$$​​
​$$a + 7\left(\frac{4a}{3}\right) = 10a + 1$$​​
 $$\Rightarrow \frac{28a}{3} = 9a + 1$$​​
28a = 27a + 3
​$$\Rightarrow \ a = 3$$​​
Now substitute a = 3 into $$d = \frac{4a}{3}:$$​​
​$$d = \frac{4 \times 3}{3} = 4$$​​
The first term of the GP is $$T_2 = 7$$​ (second term of AP), and the common ratio r = 4.
The 4th term of the GP is:
​$$T_4 = 7 \times 4^{4-1} = 7 \times 4^3 = 7 \times 64 = 448$$​​
Thus, the 4th term of the GP is 448

Thus, the correct option is (a) 448