The value of $\sum\limits_{r = 1}^n {\log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)} $ is
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
$\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^n}}}} \right)$
- C
$\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)$
- D
$\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n + 1}}}}} \right)$
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Let $a_1, a_2, a_3, \ldots$ be in an arithmetic progression of positive terms.
Let $\mathrm{A}_{\mathrm{k}}=\mathrm{a}_1{ }^2-\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2-\mathrm{a}_4{ }^2+\ldots+\mathrm{a}_{2 \mathrm{k}-1}{ }^2-\mathrm{a}_{2 \mathrm{k}}{ }^2$.
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Let $\mathrm{A}_{\mathrm{k}}=\mathrm{a}_1{ }^2-\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2-\mathrm{a}_4{ }^2+\ldots+\mathrm{a}_{2 \mathrm{k}-1}{ }^2-\mathrm{a}_{2 \mathrm{k}}{ }^2$.
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- [JEE MAIN 2024]
The income of a person is $Rs. \,3,00,000,$ in the first year and he receives an increase of $Rs.\,10,000$ to his income per year for the next $19$ years. Find the total amount, he received in $20$ years.
The income of a person is $Rs. \,3,00,000,$ in the first year and he receives an increase of $Rs.\,10,000$ to his income per year for the next $19$ years. Find the total amount, he received in $20$ years.