Question

The sum of'all natural numbers between 100 and 1000 which are multiple of 5 is :
(a) 98,450
(b) 96,450
(c) 97,450
(d) 95,450​

Answer 1

Answer:

(a)98,450

Step-by-step explanation:

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

This sequence forms an A.P.

Here,

first term, a = 105

Common difference, d = 5

Here,

a+(n−1)d = 995

=> 105+(n−1)5 = 995

=> (n−1)5 = 995−105 = 890

=> n−1 = 178

=> n = 179

Sn = n/2[2a+(n−1)d]

Sn = 179/2[2×(105)+(179−1)×(5)]

= 179/2[2(105)+(178)(5)]

= 179[105+(89)5]

= (179)[105+445]

= 179×550

= 98450

Hence, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

Answer 2

Answer:

98,450

Step-by-step explanation:

$a=105$

d=5

l=995

$now \: n th \: term \: given \: by$

$a_n=a+(n-1)d$

$995=105+5n$

$995=105+(n-1)5$

$995=100+5n-5$

$5n=995-100$

$5n=895$

$n = \frac{895}{5}$

$n=179$

Required sum

$s_n= \frac{n}{2}(a + l)$

$s_n = \frac{179}{2}(105 + 995)$

$= \frac{179}{2}(1100)$

$s_n = 98450$

$\huge \mathcal{ \fcolorbox{cyan}{black}{ \green{hope \: it \: helps}}}$