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.
Here, we have an $\mathrm{A.P.}$ with $a=3,00,000, d=10,000,$ and $n=20$ Using the sum formula, we get,
$S_{20}=\frac{20}{2}[600000+19 \times 10000]=10(790000)=79,00,000$
Hence, the person received $Rs.\, 79,00,000$ as the total amount at the end of $20$ years.
Similar Questions
Let $A =\left\{1, a _{1}, a _{2} \ldots \ldots a _{18}, 77\right\}$ be a set of integers with $1< a _{1}< a _{2}<\ldots \ldots< a _{18}<77$. Let the set $A + A =\{ x + y : x , y \in A \} \quad$ contain exactly $39$ elements. Then, the value of $a_{1}+a_{2}+\ldots \ldots+a_{18}$ is equal to...........
Let $A =\left\{1, a _{1}, a _{2} \ldots \ldots a _{18}, 77\right\}$ be a set of integers with $1< a _{1}< a _{2}<\ldots \ldots< a _{18}<77$. Let the set $A + A =\{ x + y : x , y \in A \} \quad$ contain exactly $39$ elements. Then, the value of $a_{1}+a_{2}+\ldots \ldots+a_{18}$ is equal to...........
- [JEE MAIN 2022]
A series whose $n^{th}$ term is $\left( {\frac{n}{x}} \right) + y,$ the sum of $r$ terms will be
A series whose $n^{th}$ term is $\left( {\frac{n}{x}} \right) + y,$ the sum of $r$ terms will be
The number of terms in the series $101 + 99 + 97 + ..... + 47$ is
The number of terms in the series $101 + 99 + 97 + ..... + 47$ is
The arithmetic mean of first $n$ natural number
The arithmetic mean of first $n$ natural number
If the sum of three consecutive terms of an $A.P.$ is $51$ and the product of last and first term is $273$, then the numbers are
If the sum of three consecutive terms of an $A.P.$ is $51$ and the product of last and first term is $273$, then the numbers are