The nth Term of Geometric Progression (Examples)

Example 1:

The sixth term of a geometric progression is 32 and the third term is 4. Find the first term and the common ratio.

Smart TIPS: Solving the simultaneous equation of a and r. Using the formula T_n = a r ⁿ⁻¹
Solution:

\begin{array}{l} T_{6} = 32 \\ a r^{5} = 32 ----- (1) \\ T_{3} = 4 \\ a r^{2} = 4 ----- (2) \\ \frac{(1)}{(2)} = \frac{a r^{5}}{a r^{2}} = \frac{32}{4} \\ r^{3} = 8 \\ r = 2 \\ Substitute r = 2 into (2), \\ a {(2)}^{2} = 4 \\ a = 1 \end{array}

Example 2:

In a geometric progression, the sum of the second term and the third term is 12 and the sum of the third term and the fourth term is 4, find the first term and the common ratio.

[Smart TIPS: Solving simultaneous equation to find a and r]

Solution:

T₂+ T₃= 12

ar + ar²= 12

ar (1 + r) = 12 ----- (1) ← (Factorisation)

T₃+ T₄= 4

ar² + ar³= 4

ar² (1 + r) = 4 ----- (2)

\begin{array}{l} \frac{(2)}{(1)} = \frac{a r^{2} (1 + r)}{a r (1 + r)} = \frac{4}{12} \\ r = \frac{1}{3} \\ Substitute r = \frac{1}{3} into (1), \\ a (\frac{1}{3}) (1 + \frac{1}{3}) = 12 \\ a = 27 \end{array}

\begin{array}{l}  \end{array}

Example 3

Find which term in the progression 3, 12, 48, , ... is the first to exceed 1 000 000.

[Smart TIPS: Using T_nformula for solving n]

Solution:

\begin{array}{l} 3, 12, 48, ..... \\ G P, a = 3, r = \frac{T_{2}}{T_{1}} = \frac{12}{3} = 4 \\ T_{n} > 1000000 \\ (3) {(4)}^{n - 1} > 1000000 \\ {(4)}^{n - 1} > \frac{1000000}{3} \leftarrow [(3) {(4)}^{n - 1} \neq 12^{n - 1}] \\ \log 4^{n - 1} > \log \frac{1000000}{3} \leftarrow (Put log for both sides) \\ (n - 1) \lg 4 > \lg \frac{1000000}{3} \leftarrow (\log_{a} m^{n} = n \log_{a} m) \\ (n - 1) (0.6021) > 5.523 \\ n - 1 > 9.17 \\ n > 10.17 \\ n = 11 \leftarrow (n is an integer) \\ \begin{array}{l} C h e c k : \\ T_{11} = (3) {(4)}^{10} \\ T_{11} = 3145728 > 1000000 \end{array} \end{array}