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What is the average of all multiples of $10$ from $2$ to $198$?

1. $90$
2. $100$
3. $110$
4. $120$
edited | 752 views

$a=10,l=190$

$s=\dfrac{n(a+l)}{2}=\dfrac{19\times (200)}{2}=1900$

Average $=\dfrac{1900}{19}=100$

Ans. is 100

edited
a=10

l=190

number of term=19

average of multiplae of 10= (n/2(2a+(n-1)d))/n

(2*10+18*10)/2=100

ans=100
+8
Average in case of A.P. series

average = (first + last) / 2
=(190+10)/2 = 100

10+20+30... + 180+190.

10(1+2+3+..19) Sum of first n no => n*(n+1)/2

(10*19*20)/2 = 1900.

Average = Sum / Total no = 1900/19 = (B) 100
b) 100
for finding average of any A.P. series just do
(first term+last term)/2

in this question (10+190)/2 =100
0
Good Method