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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
For a sequence $ < {a_n} > ,\;{a_1} = 2$ and $\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{1}{3}$. Then $\sum\limits_{r = 1}^{20} {{a_r}} $ is
  • A
    $\frac{{20}}{2}[4 + 19 \times 3]$
  • ✓
    $3\left( {1 - \frac{1}{{{3^{20}}}}} \right)$
  • C
    $2(1 - {3^{20}})$
  • D
    None of these
✓

Answer

Correct option: B.
$3\left( {1 - \frac{1}{{{3^{20}}}}} \right)$
b
(b) The sequence is a $G.P.$ with common ratio $\frac{1}{3}$.

Now from $\frac{{a(1 - {r^n})}}{{1 - r}},\,\,\,\,\frac{{2\,[1 - {{(1/3)}^{20}}]}}{{1 - (1/3)}}$ = $3\,\left[ {1 - \frac{1}{{{3^{20}}}}} \right]$.

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