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8. sequence and series — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS8. sequence and series1 Mark
MCQ
The value of $\sum\limits_{r = 1}^n {\log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)} $ is
  • A
    $\frac{n}{2}\log \left( {\frac{{{a^n}}}{{{b^n}}}} \right)$
  • B
    $\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^n}}}} \right)$
  • ✓
    $\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)$
  • D
    $\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n + 1}}}}} \right)$
✓

Answer

Correct option: C.
$\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)$
c
(c) The given series is

$\log a + \log \left( {\frac{{{a^2}}}{b}} \right) + \log \left( {\frac{{{a^3}}}{{{b^2}}}} \right) + \log \left( {\frac{{{a^4}}}{{{b^3}}}} \right) + ...... + \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)$

This is an $A.P.$ with first term $\log a$

and the common difference $\log \left( {\frac{{{a^2}}}{b}} \right) - \log a = \log \left( {\frac{a}{b}} \right)$

Therefore the sum of $n$ terms is

$\frac{n}{2}\left[ {\log a + \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)} \right] = \frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)$.

Trick : Check for $n = 1,\;2$.

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