QUESTION 5
Easy
Find a GP for which the sum of the first two terms is −4 and the fifth term is 4 times the third term.
SOLUTION
1
Assume the GP
Let the GP be \(a,\,ar,\,ar^2,\,ar^3,\,ar^4,...\)
2
Use the condition on the \(5^{th}\) and \(3^{rd}\) terms and solve for \(r\)
\(ar^4 = 4(ar^2)\)
Dividing by \(ar^2\), we get
\(r^2 = 4\)
⇒ \(r = 2\) or \(r = −2\)
Dividing by \(ar^2\), we get
\(r^2 = 4\)
⇒ \(r = 2\) or \(r = −2\)
3
Use the sum of first two terms
\(a + ar = −4\)
⇒ \(a(1 + r) = −4\)
⇒ \(a(1 + r) = −4\)
4
Case 1: r = 2
\(a(1 + 2) = −4\)
⇒ \(3a = −4\)
⇒ \(a = \frac{−4}{3}\)
So GP is: \(\frac{−4}{3}, \frac{−8}{3}, \frac{−16}{3},...\)
⇒ \(3a = −4\)
⇒ \(a = \frac{−4}{3}\)
So GP is: \(\frac{−4}{3}, \frac{−8}{3}, \frac{−16}{3},...\)
5
Case 2: r = −2
\(a(1 − 2) = −4\)
⇒ \(−a = −4\)
⇒ \(a = 4\)
So GP is: \(4, −8, 16, 32,...\)
⇒ \(−a = −4\)
⇒ \(a = 4\)
So GP is: \(4, −8, 16, 32,...\)
🏆
Final Answer : The GP is \(\frac{−4}{3}, \frac{−8}{3}, \frac{−16}{3},...\) or \(4, −8, 16, 32,...\)
Concept Note
For a GP, \(T_n = ar^{n−1}\)
When equations involve different terms, substitute the GP formula and solve for \(r\) first.