Question 32

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Jambmaths question: 

Given that the first and fourth term of a G.P are 6 and 162 respectively. Find the sum of the first three term of the progression

Option A: 

27

Option B: 

8

Option C: 

78

Option D: 

48

Jamb Maths Solution: 

$\begin{align}  & {{T}_{n}}=a{{r}^{n-1}} \\ & {{T}_{1}}=a=6 \\ & {{T}_{4}}=a{{r}^{3}}=162 \\ & \frac{{{T}_{4}}}{{{T}_{1}}}=\frac{a{{r}^{3}}}{a}=\frac{162}{6} \\ & {{r}^{3}}=27 \\ & r=\sqrt[3]{27} \\ & r=3 \\ & {{S}_{n}}=\frac{a({{r}^{n}}-1)}{r-1} \\ & {{S}_{3}}=\frac{6({{3}^{3}}-1)}{3-1}=3\times 26=78 \\\end{align}$

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