Which term of the ^@G.P. \space 4, 64, 1024, ....^@ upto ^@ n ^@ terms is ^@ 262144 ?^@


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Answer:

^@ 5 ^ { th } ^@

Step by Step Explanation:
  1. A geometric progression ^@(G.P.)^@ is of the form, ^@a, ar, ar^2, ar^3, ......, ^@ where ^@a^@ is called the first term and ^@r^@ is called the common ratio of the ^@G.P.^@
    The ^@n^{ th }^@ term of a ^@G.P.^@ is given by, ^@a_n= ar^{n-1} ^@
  2. Let ^@ 262144 ^@ be the ^@ n^{ th } ^@ term of the given ^@G.P.,^@ so, we need to find the value of ^@n.^@
    Here, the first term, ^@a = 4^@
    The common ratio, ^@r = \dfrac{ a_{k+1} }{ a_k } ^@ where ^@ k \ge 1 ^@
    ^@ \implies r = \dfrac{a_{1+1} }{ a_1 } = \dfrac{ a_2 }{ a_1 } = \dfrac{ 64 }{ 4 } = 16 ^@
  3. Now, @^ \begin{align} & a_{ n } = 262144 \\ \implies & ar^{ n - 1 } = 262144 \\ \implies & 4(16)^{ n-1 } = 262144 \\ \implies & 16^{ n-1 } = \dfrac{ 262144 } { 4 } \\ \implies & 16^{ n-1 } = 65536 \\ \implies & 16^{ n-1 } = 16^{ 4 } \\ \implies & n - 1 = 4 \\ \implies & n = 4 + 1 \\ \implies & n = 5 \\ \end{align}@^
  4. Hence, the ^@ 5^{th} ^@ term of the given ^@ G.P.^@ is ^@ 262144 ^@.

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