Sum of all two digit numbers which give a remainder of 4 when divided by 6 ?

English

Gujarati

Hindi

8. Sequences and Series

normal

What is the sum of all two digit numbers which give a remainder of $4$ when divided by $6$ ?

A

$777$

B

$776$

C

$780$

D

$784$

Solution

Such numbers are $10,16, \ldots . .94$

If $\mathrm{n}$ be the number of terms, then

$94=t_{n}=a+(n-1) d=10+(n-1) 6 \Rightarrow n=15$

$\therefore {\mathop{\rm sum}\nolimits} = \frac{{\rm{n}}}{2}({\rm{a}} + l) = \frac{{15}}{2}(10 + 94) = 780$

Standard 11

Mathematics

Share

Similar Questions

A manufacturer reckons that the value of a machine, which costs him $Rs.$ $15625$ will depreciate each year by $20 \% .$ Find the estimated value at the end of $5$ years.

hard

If $a,\;b,\;c$ are in $A.P.$, then $\frac{1}{{bc}},\;\frac{1}{{ca}},\;\frac{1}{{ab}}$ will be in

normal

The sum of all those terms, of the anithmetic progression $3,8,13, \ldots \ldots .373$, which are not divisible by $3$,is equal to $…….$.

hard

(JEE MAIN-2023)

The ratio of sum of $m$ and $n$ terms of an $A.P.$ is ${m^2}:{n^2}$, then the ratio of ${m^{th}}$ and ${n^{th}}$ term will be

medium

Three numbers are in $A.P.$ whose sum is $33$ and product is $792$, then the smallest number from these numbers is

easy