Question

find the sum of all 3 digit numbers which leave the remainder 2 and when divided by 5

Answer 1

hєч thєrє✌✌
ᴜʀ ᴀɴꜱ ɪꜱ ʜᴇʀᴇ ----


♥Sequence = An ordered collection of numbers a1 , a2 , a3 ,............
an......is a sequence.


♥Arithmetic progression = A sequence a1,a2,a3,.........,an is called an Arithmetic progression (A.P.) when a2-a1 = a3-a3 =......an -a(n-1) . That means A.P. is a sequence in which each term is obtained by addind a constant d to the preceding term.

♥d = The constant d is called common difference of the A.P .


♥FORMULAS -----
d= a2 - a1
an = a + (n-1 )d
sn = n/2 [2a +( n-1) d]
or
Sn = n/2 ( a + l )

where
a= first term of the A.P.
d = common difference
n = no of terms
an = nth term of A.p.
Sn= sum to nth term
l = last term


Now come to ur Question-----

ɢɪᴠᴇɴ ---
ꜱᴏ ᴛʜᴇ ꜱᴇʀɪᴇꜱ --
102 ,107 ,112,.....997


a = 102
d = 5
an = 997
n = ???
sn = ???


an = a+(n-1)d
997 = 102 + (n -1) 5
997 -102 = (n - 1) 5
895 /5 = n - 1
179 +1 = n
n = 180


Sn = n/2 ( a + l )
= 180 /2 ( 102 +997)
= 90 (1099)
= 98910


ꜱᴏ ᴛʜᴇ ꜱᴜᴍ ᴏꜰ ᴀʟʟ ᴛʜʀᴇᴇ ᴅɪɢɪᴛꜱ ɴᴜᴍʙᴇʀꜱᴡʜɪᴄʜ ʟᴇᴀᴠᴇ ᴛʜᴇ ʀᴇᴍɪɴᴅᴇʀ 2 ᴡʜᴇɴ ᴅɪᴠɪᴅᴇᴅ ʙy 3 = 98910

☺hσpє ít hєlpѕ u ☺
♥fєєl frєє tσ αѕk αnч quєrч♥