Sum of integers from 1 to 100 that are divisible by 2 or 5 is

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8. Sequences and Series

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The sum of integers from $1$ to $100$ that are divisible by $2$ or $5$ is

$3000$

$3050$

$4050$

None of these

(IIT-1984)

Solution

(b) The sum of integers from $1$ to $100$ that are divisible by $2$ or $5 =$ sum of series divisible by $2 +$ sum of series divisible by $5 -$ sum of series divisible by $2$ and $5$.

$ = (2 + 4 + 6 + …… + 100) + (5 + 10 + 15……. + 100)$

$ – (10 + 20 + 30 + …….. + 100)$

$ = \frac{{50}}{2}\left\{ {2 \times 2 + (50 – 1)2} \right\} + \frac{{20}}{2}\left\{ {2 \times 5 + (20 – 1)5} \right\}$

$ – \frac{{10}}{2}\left\{ {10 \times 2 + (10 – 1)10} \right\}$

$ = 2550 + 1050 – 550 = 3050$.

Standard 11

Mathematics

If $\tan \left(\frac{\pi}{9}\right), x, \tan \left(\frac{7 \pi}{18}\right)$ are in arithmetic progression and $\tan \left(\frac{\pi}{9}\right), y, \tan \left(\frac{5 \pi}{18}\right)$ are also in arithmetic progression, then $|x-2 y|$ is equal to:

hard

(JEE MAIN-2021)

If $3^{2 \sin 2 \alpha-1},14$ and $3^{4-2 \sin 2 \alpha}$ are the first three terms of an $A.P.$ for some $\alpha$, then the sixth term of this $A.P.$ is

hard

(JEE MAIN-2020)

If $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in $A.P.,$ prove that $a, b, c$ are in $A.P.$

hard

If $< {a_n} >$ is an $A.P$. and $a_1 + a_4 + a_7 + …….+ a_{16} = 147$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

normal

The difference between an integer and its cube is divisible by

easy

The sum of integers from $1$ to $100$ that are divisible by $2$ or $5$ is

Solution

Similar Questions

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