Given that the first and forth terms of G.P are 6 and 162 respectively, find the sum of the first three terms of the progression
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Correct Answer: Option C
Explanation:
\(a=6, U_4 = 162\\U_4 = ar^{4-1}\\U_4 = ar^3\\162 = 6(r)^3\\r^3 = 27\\r=\sqrt[3]{27}\\r=3\\S_4 = \frac{a(r^{n}-1)}{r-1}\\=\frac{6(3^{3}-1)}{3-1}\\=\frac{6(27-1)}{2}\\=3\times26\\=78\)
\(a=6, U_4 = 162\\U_4 = ar^{4-1}\\U_4 = ar^3\\162 = 6(r)^3\\r^3 = 27\\r=\sqrt[3]{27}\\r=3\\S_4 = \frac{a(r^{n}-1)}{r-1}\\=\frac{6(3^{3}-1)}{3-1}\\=\frac{6(27-1)}{2}\\=3\times26\\=78\)