Question

The 8th term of the series 256, 128, 64 ..... is​

Answer 1

Answer :

• 8th term = 2

Step-by-step explanation :

 $\underline{\underline{\bf Geometric \ Progression:}}$

•  It is the sequence of numbers such that the ratio between any two successive terms is constant.
•  In GP,

            a - first term

            r - common ratio

            n - number of terms

            Tₙ - nth term

            Sₙ - sum of n terms

• General form of G.P.,

         a , ar , ar² , ar³ ,....

• Formulae :-

         nth term of G.P.,

          $\boxed{\bf T_n=ar^{n-1}}$

         Sum of n terms,

           $\boxed{\bf S_n=\frac{a(r^n-1)}{r-1} }$

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Given series,

 256 , 128 , 64 , ...

The series is of G.P. since the ratio between successive terms is constant.

     $\bf r=\frac{128}{256} =\frac{64}{128} =\frac{1}{2}$

• first term, a = 256
• common ratio, r = 1/2

nth term is given by,

 Tₙ = arⁿ⁻¹

⇒ 8th term :

  $T_8=256 \times (\frac{1}{2})^{8-1} \\\\ T_8=256 \times (\frac{1}{2})^7 \\\\ T_8=256 \times \frac{1}{2^7} \\\\ T_8 =256 \times \frac{1}{128} \\\\ \bf T_8=2$

∴ 8th term = 2