Find the sum of all two digit numbers which when divided by $4,$ yields $1$ as remainder.
Find the sum of all two digit numbers which when divided by $4,$ yields $1$ as remainder.
The two-digit numbers, which when divided by $4,$ yield $1$ as remainder, are $13,17, \ldots 97$
This series forms an $A.P.$ with first term $13$ and common difference $4$
Let n be the number of terms of the $A.P.$
It is known that the $n^{th}$ term of an $A.P.$ is given by, $a_{n}=a+(n-1) d$
$\therefore 97=13+(n-1)(4)$
$\Rightarrow 4(n-1)=84$
$\Rightarrow n-1=21$
$\Rightarrow n=22$
Sum of n terms of an $A.P.$ is given by
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\therefore S_{22}=\frac{22}{2}[2(13)+(22-1)(4)]$
$=11[26+84]$
$=1210$
Thus, the required sum is $1210 .$
Similar Questions
If $f(x + y,x - y) = xy\,,$ then the arithmetic mean of $f(x,y)$ and $f(y,x)$ is
If $f(x + y,x - y) = xy\,,$ then the arithmetic mean of $f(x,y)$ and $f(y,x)$ is
Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=n \frac{n^{2}+5}{4}$
Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=n \frac{n^{2}+5}{4}$
There are $15$ terms in an arithmetic progression. Its first term is $5$ and their sum is $390$. The middle term is
There are $15$ terms in an arithmetic progression. Its first term is $5$ and their sum is $390$. The middle term is
Let ${a_1},{a_2},.......,{a_{30}}$ be an $A.P.$, $S = \sum\limits_{i = 1}^{30} {{a_i}} $ and $T = \sum\limits_{i = 1}^{15} {{a_{2i - 1}}} $.If ${a_5} = 27$ and $S - 2T = 75$ , then $a_{10}$ is equal to
Let ${a_1},{a_2},.......,{a_{30}}$ be an $A.P.$, $S = \sum\limits_{i = 1}^{30} {{a_i}} $ and $T = \sum\limits_{i = 1}^{15} {{a_{2i - 1}}} $.If ${a_5} = 27$ and $S - 2T = 75$ , then $a_{10}$ is equal to
- [JEE MAIN 2019]
Let $a_1, a_2, \ldots \ldots, a_n$ be in A.P. If $a_5=2 a_3$ and $a_{11}=18$, then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots . \cdot \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to $..........$.
Let $a_1, a_2, \ldots \ldots, a_n$ be in A.P. If $a_5=2 a_3$ and $a_{11}=18$, then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots . \cdot \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to $..........$.
- [JEE MAIN 2023]