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

  1. $90$
  2. $100$
  3. $110$
  4. $120$
in Numerical Ability by Veteran (105k points)
edited by | 986 views

5 Answers

+15 votes
Best answer

$ a=10,l=190$

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

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

Ans. is 100

Correct Answer: $B$

by (339 points)
edited by
+2
It's an AP $10,20,.......190$

The average of an AP=$\frac{a_1+a_n}{2}=\frac{190+10}{2}$
+1

Average in case of A.P. series

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

+7 votes
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
by (143 points)
+11
Average in case of A.P. series
 
average = (first + last) / 2
        =(190+10)/2 = 100
+5 votes
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
by Boss (41.5k points)
0 votes
b) 100
by Loyal (5.2k points)
0 votes
for finding average of any A.P. series just do
      (first term+last term)/2

in this question (10+190)/2 =100
by Junior (937 points)
0
Good Method
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