QUESTION 1
Easy
Find the \(12^{th}\) term of a GP with common ratio 2, whose \(8^{th}\) term is 192.
SOLUTION
1
Note down what is given
Common ratio, \(r = 2\)
\(8^{th}\) term, \(T_8 = 192\)
\(8^{th}\) term, \(T_8 = 192\)
2
Find the first term
The GP formula is: \(T_n = ar^{n − 1}\)
We have,
\(T_8 = 192\)
⇒ \(a(2)^{8 − 1} = 192\)
⇒ \(a(128) = 192\)
⇒ \(a = \frac{192}{128}\)
⇒ \(a = \frac{3}{2}\)
We have,
\(T_8 = 192\)
⇒ \(a(2)^{8 − 1} = 192\)
⇒ \(a(128) = 192\)
⇒ \(a = \frac{192}{128}\)
⇒ \(a = \frac{3}{2}\)
3
Find the \(12^{th}\) term
\(\begin{aligned} T_{12} &= ar^{12 − 1} \\ &= \frac{3}{2} × 2048 \\ &= 3072 \end{aligned}\)
🏆
Final Answer : \(T_{12} = 3072\)
Concept Note
A geometric progression(GP) is a sequence in which each term is obtained by multiplying the previous term by a fixed number called the common ratio(r). Its formula is:
\(T_n = ar^{n − 1}\)
where \(a\) is the first term and \(r\) is the common ratio.