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Mathematics

The 5th term from the end of the G.P. 2, 6, 18, …., 13122 is

  1. 162

  2. 486

  3. 54

  4. 1458

AP GP

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Answer

The above series is a G.P. with a = 2 and d = 62=3.\dfrac{6}{2} = 3.

nth from end = l(1r)n1l\Big(\dfrac{1}{r}\Big)^{n - 1}

∴ 5th term from end = 13122×(13)5113122 \times \Big(\dfrac{1}{3}\Big)^{5 - 1}

=13122×134=1312281=162.= 13122 \times \dfrac{1}{3^4} \\[1em] = \dfrac{13122}{81} \\[1em] = 162.

Hence, Option 1 is the correct option.

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