The value of -i-1nlogaT-br-1A- n-2logan-bnB- n-2logan-1-bn-1C- n-2logan-1-bnD- n-2logan-1-bn-1

The correct option is B The given series is loga+log(a^2/b)+log(a^3/b^2)+log(a^4/b^3)+............+log(a^n/b^n 1) This is an A.P. with first term loga and the ...

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Byju's Answer

Standard XI

Mathematics

Sum of n Terms in AP

The value of ...

Question

The value of $n \sum r = 1 l o g (\frac{a^{r}}{b^{r - 1}})$ is

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Solution

The correct option is C
The given series is
$l o g a + l o g (\frac{a^{2}}{b}) + l o g (\frac{a^{3}}{b^{2}}) + l o g (\frac{a^{4}}{b^{3}}) + . . . . . . . . . . . . + l o g (\frac{a^{n}}{b^{n - 1}})$
This is an A.P. with first term loga and the common difference
$l o g (\frac{a^{2}}{b}) - l o g a = l o g (\frac{a}{b})$
Therefore the sum of n terms is
$\frac{n}{2} [l o g a + l o g (\frac{a^{n}}{b^{n - 1}})] = \frac{n}{2} l o g (\frac{a^{n + 1}}{b^{n - 1}})$
Trick : Check for $n = 1, 2.$

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The value of n∑r=1log(arbr−1) is

The correct option is C The given series is loga+log(a2b)+log(a3b2)+log(a4b3)+............+log(anbn−1) This is an A.P. with first term loga and the common difference log(a2b)−loga=log(ab) Therefore the sum of n terms is n2 [ loga+log(anbn−1)] = n2log (an+1bn−1) ​ Trick : Check for n=1,2.

The value of $n \sum r = 1 l o g (\frac{a^{r}}{b^{r - 1}})$ is