Determine the third term of the G.P, Whose common ratio is 3 and the sum to first 7 terms is 2186?

Correct option is B

Given:

Common ratio is 3 and the sum to first 7 terms is 2186.

Formula Used:

​$$S_n = \frac{a(r^n - 1)}{r - 1}$$

​$$a_n = a \cdot r^{n-1}$$​​

where:
​$$S_n$$​ is the sum of the first n terms,
a is the first term,
r is the common ratio,
n is the number of terms.

Solution:

r = 3

$$S_7 = 2186$$

​n = 7

​$$S_n = \frac{a(r^n - 1)}{r - 1}$$

​$$​​2186 = \frac{a(3^7 - 1)}{3 - 1}$$​​

​$$2186 = \frac{a(2187 - 1)}{2}$$​​

​$$2186 = \frac{a(2186)}{2}$$​​

​$$2186 \times 2 = a \times 2186$$​​

​$$4372 = a \times 2186$$

$$a = \frac{4372}{2186} = 2$$

So, the first term a = 2.​​​

​$$a_n = a \cdot r^{n-1}$$

$$a_3 = 2 \times 3^{3-1}$$

$$a_3= 2 \times 3^2$$

$$a_3= 2 \times 9 = 18$$

So, the third term of the GP is 18.

Thus, the correct answer is (b).