What is the average of first 100 multiples of 5.

M4maths Previous Puzzles

19 Aug 2014 Find the average What is the average of first 100 multiples of 5.: OPtion
1. 125
2. 55
3. 152
4. 154.5
5. 175
6. 235
7. 252.5
8. 255.5
9. 275.5
10. None of these; Correct Option: 7 Best Solution (205)

Navneet Kaur 10 years AGO

sum of n natural nos.=n(n+1)/2
and we are given
5+10+15+20+25+.....+500
5(1+2+3+4+....+100.)
so sum=5(100*101)/2
sum=25250
average=sum/100
so average=252.50

Like? Yes (10) | No

ABHISHEK 10 years AGO

avg=sum/n={n/2(2a+(n-1)d)}/n
so,avg={100/2(2*5+99*5)}/100
=252.5

Like? Yes (9) | No (2)

KAUSHAL KUMAR 10 years AGO

sum of 1st 100 multiples of 5 are (100/2)*(5+500)=25250
avg=25250/100=252.5.

Like? Yes (3) | No

MRITUNJAY KUMAR 10 years AGO

5+500/2=252.5
option 7)

Like? Yes (4) | No (1)

Grishma Shah 10 years AGO

Multiple of 5 are 5,10,15,20,25....
Converting to A.P.

a=5, d=5, n=100

Sn= n/2(2a+(n-1)d)
= 100/2 (2*5 + (100-1)5)
=25250
The average of this 100 multiple are
25250/100 = 252.5

Answer is option 7

Like? Yes (3) | No

RAKESH 10 years AGO

5*(1+2+3+.....+100)/100 = 252.5

Like? Yes (2) | No

ABHIJIT DUTTA 10 years AGO

5x1 + 5x2 + 5x3 + 5x4 ... 5x100 =
5 * (1+2+3+4+ ... 100) =
5 * (50.5 x 100) =
5 * 5050 =
25,250

average = 25250/100 = 252.5

option 7)

Like? Yes (2) | No

LOKESH SAMANT 10 years AGO

What is the average of first 100 multiples of

Like? Yes (2) | No

Harshad 10 years AGO

5+10+15+20+....500=25250
Average=25250/100=252.5

Like? Yes (2) | No

Kushagra Tiwari 10 years AGO

first 100 multiple of 100 =>>>
5x1 + 5x2 + 5x3 + 5x4 ... 5x100 =
5x(1+2+3+4+ ... 100) =
5x(50.5 x 100) =
5x5050 =
25,250

So avrage multiple of 100 = 25,250/100
= 252.5

Like? Yes (4) | No (2)

KRISHNA CHANDRA UPADHYAY 10 years AGO

5x1 + 5x2 + 5x3 + 5x4 ... 5x100 =
5x(1+2+3+4+ ... 100) =
5x(50.5 x 100) =
5x5050 =
25,250
Therefore, avg.=25250/100=252.50.
HENCE OPTION (7) is correct.

Like? Yes (2) | No

Sourojit Dhua 10 years AGO

5x(1+2+....+100)=avg. x 100
so,
avg. = 252.5

Like? Yes (2) | No

Geetha Penigalapati 10 years AGO

Required average = 5(1+2+3+.........+100)/100
= 5(100(101)/2)100 = 252.5
since sum of first n natural numbers =n(n+1)/2

Like? Yes (2) | No

Vinayak Chaskar 10 years AGO

5+10+15+20+25+..............;
5*1 + 5*2 + 5*3 + 5*4 +..............+5*100;
5*(1+2+3+4+5+........+100);
(5*100*101)/(2) = 25250;
avg of number = 25250/100 = 252.5;

Like? Yes (2) | No

pratibha kushwaha 10 years AGO

first 100 multiple of 5
5+10+-----500
5(1+2+3---+100)
5(100*(101))/2
average of 100 multiple of 5
5(100*101)/(2*100)
252.5

Like? Yes (1) | No

HARDIP BHALANI 10 years AGO

5+10+15+20+....500=25250
Average=25250/100=252.5

Like? Yes (2) | No (1)

Sneh Siddhi 10 years AGO

sum=100/2(500+5)
avg=sum/100

Like? Yes (1) | No

P.Someshwar Reddy 10 years AGO

the 1st multiple of 5 is 5 => a=5
100th multiple of 5 is 500 => l=500
number of terms =100 = > n=100
sum of n terms is n/2(a+l)
=> 100/2(5+500)
=>25250
and to get average of 100 divide sum by 100
=>252500/100
=>252.5
Option (7)

Like? Yes (1) | No

Ravi Kumar 10 years AGO

sum of ap/100=
( S_n={n}/{2}[ 2a_1 + (n-1)d])/100
252.5

Like? Yes (1) | No

Gourab Dutta 10 years AGO

Sum of the multiples=n/2(2a+(n-1)*d)
=100/2(2*5+(100-1)*5)=505*50
Avg=505*50/2=252.5

Like? Yes (2) | No (1)

Saraswathy 10 years AGO

5 + 10 +First 100 multiples of 5 ll be
5, 10, 15, 20, ....., 500
a=5, n=100, l=500
Sn = n/2 * (a+l)
Sn = 100/2 * (5+500)
Sn = 25250
Average ll be 25250/100 = 252.5

Ans : 252.5 (Option 7)

Like? Yes (1) | No

Ajay Kumar Gupta 10 years AGO

sum of 100 multiples of 5=100*(5+500)/2=25250
average=25250/100=252.5

Like? Yes (1) | No

Abhishek Vishwakarma 10 years AGO

5x1 + 5x2 + 5x3 + 5x4 ... 5x100/100 =
5x(1+2+3+4+ ... 100)/100 =
5x(50.5 x 100)/100 =
5x5050/100 =
25,250/100
252.5

Like? Yes (1) | No

PAWAN KUMAR 10 years AGO

we could see that when we increase the number of multiples of five the average is increased by 2.5.
therefore,
the average of first 100 multiples of five = 5+99*2.5
= 5+247.5
= 252.5 Ans.

Like? Yes (1) | No

Abhinav khare 10 years AGO

sum of an AP S= n/2 [2a+(n-1)d]
so, n=100
a=5
d=5
on solving we get 252.5

Like? Yes (1) | No

suman kumar 10 years AGO

sum of all first 100 mutiples of 5 taken=100/2(5+500). due to sum of all terms in arithmetic progression.its average will be whole divided by 100. which is equal to 252.5 or u may say since difference is same so avg.=(final term +initial term)/2 which is the same 252.5

Like? Yes (1) | No

Devendra Marghade 10 years AGO

average= (Sum of 100 multiples of 5)/ 100
=(100/2)[2*5 + 99*5]/ 100
=505/2
=252.5

Like? Yes (1) | No

AMRENDRA KUMAR 10 years AGO

100 multiple of 5 is 500
so,5+10+15and so on=505*50
average=505*50/100=252.5

Like? Yes (1) | No

Sneha 10 years AGO

5(1+2+3+....+100)
[Formula: Sum of n natural numbers=n(n+1)/2]
=5((100*101)/2)
=5(50*101)
=25250
Average=25250/100=252.5

Like? Yes (1) | No

Yogesh 10 years AGO

sequence is in A.P.
5,10,15,20,25.......490,495,500

sum=100/2*(2*5+(100-1)*5)
=50*505
=25250
Sum/100 to find the average
=25250/100
=252.5

Like? Yes (1) | No

Varunraj Reddy 10 years AGO

5*(1+2+3+...+100)/100
=5*(100*101/2)/100
=5*5050/100
=252.5

Option 7)

Like? Yes (1) | No

Md.Sultan Khichi 10 years AGO

Fsrt term=5, did=5
Last term=5+(n-1)5=500
Sum=n/2(a+last term)=25250
Average=25250/100=252.50
Ans on.7

Like? Yes (1) | No

Swechchha Agarwal 10 years AGO

5,10,15,...100 terms
its an a.p. so sum=n/2(2a+(n-1)d)=100/2(2*5+(100-1)5)=25250
average=sum/100=252.50

option 7)

Like? Yes (1) | No

nigarashraf 10 years AGO

5(1+2+3+........100)/100=252.5

Like? Yes (1) | No

praveenkumar 10 years AGO

5x1 + 5x2 + 5x3 + 5x4 ... 5x100 =
5x(1+2+3+4+ ... 100) =
5x(50.5 x 100) =
5x5050 =
25,250

Like? Yes (1) | No

Nilanjan Mondal 10 years AGO

in AP series
1st no = 5,
diff = 5
total numbers =100
sum = (100(2*5-((100-1)*5)))/2=25250
avg = 25250/100 = 252.5

Like? Yes (1) | No

RAJEEV KUMAR 10 years AGO

5x1 + 5x2 + 5x3 + 5x4 ... 5x100 =
5x(1+2+3+4+ ... 100) =
5x(50.5 x 100) =
5x5050 =
25,250/100=252.5

Like? Yes (1) | No

Rahul rahul 10 years AGO

this form an AP -) 5,10,15,20.......500

total terms= a+(n-1)d , where a=5,d=common

difference=5

so we got n=500

now sum of the above AP is n/2[2a+(n-1)d]

100/2[2*5+(100-1)5]=2525
avg.=2525/100=252.5

Like? Yes (1) | No

AMIT SINGH 10 years AGO

sum of 1st 100 multiples of 5 are (100/2)*(5+500)=25250
avg=25250/100=252.5.

Like? Yes (1) | No

Ashutosh Bidaliya 10 years AGO

(5+500)/2=252.5

Like? Yes (1) | No

ravi shah 10 years AGO

it is an a.p series. sum of 1st 100 multiples of 5=n/2[2a+(n-1)d]=100/2[2*5+99*5]=25250
so, average=25250/100=252.5

Like? Yes (1) | No

Pranay Mukhopadhyay 10 years AGO

sum of 1st 100 multiples of 5=100/2[2*5+(100-1)*5]=25250
avg of this sum=25250/100=252.5

Like? Yes (1) | No

LOGAN WOLVERINE 10 years AGO

x(n+1)/2=5(100+1)/2=252.5

Like? Yes (1) | No

Mou Chakraborty 10 years AGO

According to the a.p. series,the sum of 100 mutipules of 5 is 25250.
100/2(10+(99*5))=25250
then,the average will be 25250/100=252.5

Like? Yes (1) | No

Akshay V K 10 years AGO

5+10+........100 terms
sum=25250
avg=sum/total no of terms
avg=252.5

Like? Yes (1) | No

sompon temtheerakij 10 years AGO

sum of first 100 multiples of 5 = 5+10+15+20+25+30+35+40+45+50+.....490+495+500 = 25250
The average of first 100 multiples of 5 = 25250/100 = 252.5

Like? Yes (1) | No

Vivek 10 years AGO

L=5+(100-5)5=500
Avg would be avg of 1st and last=
(5+500)÷2=252.5

Like? Yes (1) | No

Urvashi Jain 10 years AGO

tn=a+(n-1)xd
5,10,15........500
500=5+(n-1)x5
n=100
sn=n/2[2a+(n-1)xd]
sn=100/2[2x5+(100-1)x5]
sn=50x505
according to the question we find the avg
so divide by 100
sn=25250
25250/100
252.5
option(7)

Like? Yes (1) | No

yagnik pandya 10 years AGO

5, 10, 15, 20, 25,...
So average is 252.5
Hence option 7

Like? Yes (1) | No

Jagadeeswar Reddy.J 10 years AGO

A.P for first 100 multiples of 5 is
Sum = 100/2(2*5+(100-1)5)
= 50500/2
= 25250
So average of first 100 multiples of 5 = 25250/100
= 252.5

Like? Yes (1) | No

Dark Knight 10 years AGO

Ap question
{n/2(first term+last term)}/100
hence
{100/2(5+500)}/100(since avg)
hence answer is 252.5

Like? Yes (1) | No

swarupa nuggu 10 years AGO

252.5

Like? Yes (1) | No

rutuja 10 years AGO

(5*1+5*2+.....+5*100)/100=5(1+2+....+100)/100
=5*(100*101)/100*2
=25250/100
=252.50

Like? Yes (1) | No

padmaja 10 years AGO

(5*(1+2+..........+100))/100
=(5*((100*101)/2))/100
=252.5

Like? Yes (1) | No

Surit Datta 10 years AGO

here summation of 1st 100 multiple of 5 is
5(1+2+3+4+........+100)
which is (5*100*101)/2=25250
according to the formula n(n+1)/2
multiplied by 5 as multiple of 5
so average of the summation is 25250/100
which is 252.50
thats it...

Like? Yes (1) | No

Partha Pratim Mahata 10 years AGO

5*(1+2+3+4......+100)=5*101*50=25250
so avg is 25250/100=252.5
so option 7 is correct

Like? Yes (1) | No

vikash kumar 10 years AGO

the average of first 100 multiples of 5 is the average of 50th multiple and 51th multiple=(5*50+5*51)/2=252.5

Like? Yes (1) | No

Abhinav 10 years AGO

find the sum of first 100 natural no. then multiply it by 5 and then devide it by 100...... you have your answer 252.5

Like? Yes (1) | No

shree 10 years AGO

252.5

Like? Yes (1) | No

prashant kumar 10 years AGO

sum off all no=100/2(2*5+(100-1)*5)
so avg=sum of all no./100=50*5(2+99)/100=252.5

Like? Yes (1) | No

Shikha 10 years AGO

avg of 100 multiples of 5=5(1+2+......+100)/100
=(5*100*101)/(2*100)=505/2=252.5

Like? Yes (1) | No

sharathbabu 10 years AGO

Required average =5(1+2+3+…….+100)/100 => (5*100*101)/(100*2) => (505/2)=252.5

Like? Yes (1) | No

sandeep yendamuri 10 years AGO

i wont tell you... :P

Like? Yes (1) | No

indumathi 10 years AGO

5(1+2....+100)=5(n(n+1)/2)

Like? Yes (1) | No

Dinesh thatti 10 years AGO

5x1 + 5x2 + 5x3 + 5x4 ... 5x100 =
5x(1+2+3+4+ ... 100) =
5x(50.5 x 100) =
5x5050 =
25,250

we want average so 25250/100=252.5

Like? Yes (1) | No

Indira 10 years AGO

first 100 multiples of 5 = 5+10+.......+500
avg =(5+10+15.....+500)/100
= (5(1+2+3........+100)/100)
=5((100*101)/2)*100
=252.5

Like? Yes (1) | No

Pallab Das 10 years AGO

5x(1+2+3+4+ ... 100)
5x5050
25250
avg=252.5

Like? Yes (1) | No

T.VIGNESWARA REDDY 10 years AGO

5 multiples are 5,10,15,----------500 sum of all terms is 100/2(5+500) is 25250 so average is 252.5

Like? Yes (1) | No

Alok Maurya 10 years AGO

5+10+15+ ...............................+495+500
Sum of n term=(n/2)(2a+(n-1)d)
=>100/2[2*5+99*5]
=>50*505
=>252.50
now,
average=25250/100=252.50

Like? Yes (1) | No

Smriti Priya 10 years AGO

sum of first 100 multiples of 5=5+10+15+...
=5(1+2+3+..+100)
=5*100*101/2
average=(5*100*101)/(100*2)
=5*101/2
=252.5

Like? Yes (1) | No

ANKAJ KUMAR SINGH 10 years AGO

tn= a+(n-1)xd
= 5+(100-1)*5
=500
sum= n/2(a+l)
=50x(505)
25250=
avg=25250/100
=252.5

Like? Yes (1) | No

PRAVIN KUMAR 10 years AGO

5(1+2+.........................+100)/100
so avg=252.5

Like? Yes (1) | No

anandhakumar 10 years AGO

5x1 + 5x2 + 5x3 + 5x4 ... 5x100
5x(1+2+3+4+ ... 100)
5x(50.5 x 100)
5x5050
25,250/100=252.5
option 7

Like? Yes (1) | No

nikita 10 years AGO

(5(1+2+3+4.....+99+100))/100
=5*(100(100+1))/(2*100)
=252.5

Like? Yes (1) | No

Manu K N 10 years AGO

MIDDLE NOS OF 5 AND 500 ARE 250 AND 255. SO THEIR AVERAGE IS 252.5

Like? Yes (1) | No

S.SHAFI 10 years AGO

5(1+2+...100)/100 = 252.5

Like? Yes (1) | No

rakesh kumar 10 years AGO

Avg.or Am=(5+500)/2=252.5

Like? Yes (1) | No

rohit sakhi 10 years AGO

Sum=100\2(2*5+(100-1)*5)
Avrg=sum\100

Like? Yes (1) | No

SUNITEE GARIMA 10 years AGO

the series is 5,10,15,..........................upto 100 terms
first term=5,common difference=5,no. of terms=100
sum upto 100 terms=100/2[2*5+(100-1)5]
average upto 100 terms=sum/100=252.5

Like? Yes (1) | No

Ashish Kumar gupta 10 years AGO

5,10,15,20....... upto100th term

S=(100/2)(2*5+(100-1)*5)
=50(10+495)
sum of the series=25250

avg=(sum of the series/number of terms)
=(25250/100)
=252.50

Like? Yes (1) | No

ANKUR ANAND 10 years AGO

[100*(5+500)/2]/100=252.5

Like? Yes (1) | No

B.Venkat Reddy 10 years AGO

Since average of n multiples =a(n*(n+1))/2*n ;where a is the starting term of multiple
sum of 100 multiples of 5 = 5+10+15...+500
=5(1+2+3+....+100)
=5(100*101)/2
average of 100 multiples=5(100*101)/(2*100)
=505/2
=252.5

Like? Yes (1) | No

RITESH SINGH 10 years AGO

First 100 multiples of 5 are-5*1,5*2,5*3,.........5*100
therefore avg=(5*1+5*2+5*3+......5*100)/100
=5(1+2+3+.......100)/100
=5*100*(100+1)/100
=252.5

Like? Yes (1) | No

Ashok Reddy Kalam 10 years AGO

5(1+2+3+....+100)/100=5*100*101/(2*100)=252.5

Like? Yes (1) | No

sudhakar 10 years AGO

The answer is option 7)252.5

Like? Yes (1) | No

selvakumar 10 years AGO

option 7)252.5

Like? Yes (1) | No

ganesh babu 10 years AGO

5+10+15+....+500=5(1+2+...+100)=5(100)101/2{n(n+1)/2}
average=5x100x101/200=252.5

Like? Yes (1) | No

Mukesh Sharma 10 years AGO

We have to find 5*(1+2+...100)/100
Formula for finding sum of n integers = n*(n+1)/2
Lets find out sum of 1 to 100 = 100*101/2= 10100/2= 5050
Then 5*5050/100=5*50.5=252.5

Like? Yes (1) | No

jagdish sharma 10 years AGO

5+10+15.................upto 100times
sum=n/2[2*a+(n-1)d]
n=100;a=5;d=5
sum=25250
avg=25250/100=252.50

Like? Yes (1) | No

kumar mayank 10 years AGO

average= total sum/100
(100*(5+500))/200
=252.5

Like? Yes (1) | No

prem kumar 10 years AGO

5(1+2+3+.......100)/100
=5*5050/100
=252.5

Like? Yes (1) | No

surya 10 years AGO

5(1+2+3+..........+100)/100=(5(100*101)/2)/100=505/2=252.5

Like? Yes (1) | No

krishna kanth 10 years AGO

5+10+.........500/100
5(1+2+3.......100)/100
5*100*101\()100*2)=252.5

Like? Yes (1) | No

LAKSHMIKANTH 10 years AGO

(50*(10+(99*5))/100=252.5

Like? Yes (1) | No

kalpana kumari 10 years AGO

a.p series n(n+1)/2

Like? Yes (1) | No

sundeep gullapudi 10 years AGO

5(1+2+3...+100)/100 = 252.5

Like? Yes (1) | No

Vanika 10 years AGO

using arithmetic progression 100th multiple will be
=500
sum of first 100 multipleof 5 = 25250
so average= 252.5

Like? Yes (1) | No

Vignesh 10 years AGO

100 multiples of 5 are as follows,
5,10,15,20,25,30,35,40,...............500

so (5+500)/2 = 505/2
= 252.5

option 7

Like? Yes (1) | No

gangulasiva 10 years AGO

252.5
take first multiple 5 and last miltiple 500
theaverage is 252.5

Like? Yes (1) | No

avinash 10 years AGO

(5+500)/2
1st mul of 5 is 5 and 100 mul of 5 is 500
avg of both is 252.5

Like? Yes (1) | No

Manish Singhal 10 years AGO

multiple of 5
5 10 15 ...
5(1 2 3 4 5.... 100) // 100 multiple of 5
sum of 1... 100
n(n+1)/2= 100(100+1)/2=50*101
average of multiple of 5 =5*50*101/100=252.5

Like? Yes (1) | No

sudipta maji 10 years AGO

(5x1 + 5x2 + 5x3 + 5x4 ... 5x100 )/100= 252.5

Like? Yes (1) | No

nisha sharma 10 years AGO

Using AP,
sum of n terms of AP is: n/2(2a+(n-1)d)
here we have to find sum of all term & when we divide it by total no of term then we get average
therefore here,
no of term is 100
common difference=5
first term will be=5
hence sum of 100 term will be 50(2*5+(100-1)*5)=25250
average=25250/100=252.5

Like? Yes (1) | No

abhishek 10 years AGO

sum of n term=n/2[2a+(n-1)d]
=100/2[(2*5)+(100-1)5]
=25250.
avg=25250/100
=252.5

Like? Yes (1) | No

GOWRABATHUNI RAJESH 10 years AGO

100 multiples of 5 is

5(100)(101)/100*2=252.5

Like? Yes (1) | No

Shaban Ahmad 10 years AGO

this will make an AP with 1st term=5, last term=500 and common difference=5
so the average of 1st and last term will be the average of AP
hence, the average=5+500/2=252.5

Like? Yes (1) | No

joydeep adhikary 10 years AGO

5(1+2+3+....+100) as 1+2+3+..n =1/2*n(n+1)
5*0.5(100)(100+1)=25250
now div by 100 as avg is reqd..

Like? Yes (1) | No

mukul kumar bhardwaj 10 years AGO

first 100 multiples of 5 will be 5(1+2+3...+100)=5(100*101)/2 and for average we divide it by 100 to get 252.5

Like? Yes (1) | No

Deepak Kumar 10 years AGO

5 x (1+2+3+4+5 ...100)/100= 252.5

Like? Yes (1) | No

T.RAJAMADHANGI 10 years AGO

Multiples of 5 so 5 to 500.
5+10+15+20+25+30+35+40+45+50+55+60+65+70+75+80+85+90+95=950
similarly 100+105+.....................+195=2950
200+205+...................+295=4950
300+305+.................+395=6950
400+405+.................+495=8950
100*5=500
then add all the things 950+2950+4950+6950+8950+500=25250
then the average is 25250/100=252.5
Option 7.

Like? Yes (1) | No

SHANMUGAPRIIYAN 10 years AGO

5+10...............95=950
100+105..........195=2950
200+205...........295=4950
300+305............395=6950
400+405............495=8950
5*100=500
Add all the things950+2950+4950+6950+8950+500=25250
average is 25250/100=252.5
Option 7

Like? Yes (1) | No

deepesh singh panwar 10 years AGO

this question is related to arthmatic procession

then sn=n/2{2a+(n-1)d}
according to the question,
a=5,d=5,n=100
s100= 100/2{2*5+(100-1)5}
=50{10+(99*5)
50{10+495}
=50*505=25250
avg=25250/100
=252.50

Like? Yes (1) | No

kajol nigam 10 years AGO

5+10+15+20.......500=(100/2)(2*5+(100-1)5)
=25250

avg=25250/100
= 252.5

Like? Yes (1) | No

Pranoy Bej 10 years AGO

1st multiple of 5 is 5 and 100th multiple of 5 is 500. so
avg = (5+500)/2 = 252.5

Like? Yes (1) | No

Venkateswarlu 10 years AGO

5 multiples series=5,10,15,20,-- - - - - - - ,100 numbers
It is an AP,,,, a=5,d=5 n=100
S=(n/2)*(2a+(n-1)*d)
S=25250
avg=25250/100=252.5

Like? Yes (1) | No

santu ghosh 10 years AGO

the average of first 100=((100*(100+1))/(2*100)=50.5
the ans is =50.5*5=252.5

Like? Yes (1) | No

Koushik mondal 10 years AGO

(5+10+...+500)/100
=5(1+2+...+100)/100
=5*100*101/100

Like? Yes (1) | No

RAVI RANJAN 10 years AGO

5(1+2+3+..........+100)/100=5*100*101/100*2=252.5

Like? Yes (1) | No

SHASHANK VERMA 10 years AGO

multiple of 5 =5,10,15,20,.....

a=5;d=5;n=100;

sum=(n/2)(2a+(n-1)d)

sum=50*(10+99*5)
average=sum/100
=50*(10+99*5)/100
=252.5

Like? Yes (1) | No

Radhika Rathore 10 years AGO

as first 100 multiple of 5 are 5,10,15.....
which is an A.P so we can found their sum bu formula of sum of n series in A.p i.e n/2(2a+(n-1)d)
=> 100/2(2*5+(100-1)*5)
=>100(252.5)
there for avg is : sum/100
=> 100(252.5)/100
=>252.5(answer)

Like? Yes (1) | No

rangababu gudimetla 10 years AGO

by using (n(n+1))/2,here n=100..ans=(5(n(n+1))/2)100=252.5

Like? Yes (1) | No

srinivas 10 years AGO

avg of (1,2,.......,100)*5=(((100*101)/2)*5)/100

Like? Yes (1) | No

rakesh yadav 10 years AGO

avg=(100/2)(10+99*5)/100->252.5

Like? Yes (1) | No

bhanu saileela 10 years AGO

sum of 5+10+15+.....+500
=5(1+2+3+....100)
=25250
average=25250/100=252.50

Like? Yes (1) | No

sonakshi gupta 10 years AGO

252.5

Like? Yes (1) | No

raviteja 10 years AGO

This would be a series: 5 + 10 + 15 + 20 +...+ 500
there are n terms → n = 100
the nth term is 500

to find the sum of the series, you take the sum of the first and last terms and multiply it by (n/2)

(5 + 500) * (100 / 2)

505 * 50 = 25250

finally, 25250/100= 252.5

Like? Yes (1) | No

raghuram dharmavarpu 10 years AGO

505/2
formulae n/2(2a+(n-1)d)
l=a+(n-1)d

Like? Yes (1) | No

Tark Patel 10 years AGO

Average = ( (5*1) + (5*2) + (5*3) + ..... + (5*100) ) / 100
= ( 5 * ( 1 + 2 + 3 + ..... + 100 ) ) / 100
= ( 5 * ( (100 * ( 100 + 1 ) ) / 2 ) ) / 100
= ( 5050 * 5 ) / 100
= 252.5

Like? Yes (1) | No

SUREKHA 10 years AGO

work it out... :)

Like? Yes (1) | No

roshan choudhury 10 years AGO

An=a+(n-1)d
=5+99*5=500
500 is the last no
sum of no is=(500+5)*100/2
25250
average =25250/100=252.5

Like? Yes (1) | No

Jasmita Parmar 10 years AGO

5 * 1 + 5 * 2 + 5 * 3 + 5 * 4 ... 5 * 100 =
5 * (1+2+3+4+ ... 100) =
5 * ( (100*101)/2) ) =
5 * 5050 =
25,250

Avg = 25250/100 =252.50

Like? Yes (1) | No

mrunali s gaonkar 10 years AGO

{5x1 + 5x2 + 5x3 + 5x4 ... 5x100 }/100=
{5x(1+2+3+4+ ... 100)}/100 =
(5x{(100*101)/2})/100 =
5x5050/100 =
25,250/100=252.5

Like? Yes (1) | No

TINKU MITTAL 10 years AGO

first 100 multiples of 5 are
5,10,15,20,..............,495,500
average=sum/ no. of terms
so Sum=(100/2)*(5+500)=25250 (by A.P.)
so average=25250/100= 252.50

Like? Yes (1) | No

Aravinth kumar 10 years AGO

5,10,15......500
sum of all number = n(first number + last number)/2
=100(505)/2
=25250
avg=25250/100
ans=252.5

Like? Yes (1) | No

Samrat Adhikari 10 years AGO

First 100 multiples of 5= 5,10,15,20,........,500
Avg=(5+10+15+........+500)/100=252.5

Like? Yes (1) | No

k.dinesh kumar 10 years AGO

5x1 + 5x2 + 5x3 + 5x4 ... 5x100 =
5x(1+2+3+4+ ... 100) =
5x(50.5 x 100) =
5x5050 =
25,250
i.e 25250/100=252.5

Like? Yes (1) | No

Rashmi bharti 10 years AGO

25250/100=252.5

Like? Yes (1) | No

Kirankumar s kurubar 10 years AGO

5x1 + 5x2 + 5x3 + 5x4 ..... 5x100 = 5 * (1 + 2 + 3 + 5 .... + 100)= 5 * (sum of n numbers)For calculating the sum of n numbers we have the formulai.e (n * (n + 1)) / 2apply the formula in the above equation.where n = 100so= 5 * (100 * (100 + 1)) / 2= 5 * (50 * 101)= 5 * 50 * 101= 25250 This the sum of the first 100 positive number which is multiples of 5.And for average:= 25250 /100= 252.50

Like? Yes (1) | No

pavankumar 10 years AGO

first 100 multiples of 5 is 5(1+2+3+......+100)=n(n+1)/2
5*n(n+1)/2*100=252.2

Like? Yes (1) | No

sruthi 10 years AGO

every 10th multiple of 5 ends as 50,100,150,200,250,300,350,400,450,500.
so up to 100th multiple is 500
now 5+10+15+20...+500
5(1+2+3+...+100)=5(n(n+1)/2)
==>5*((100*101)/2)=5050*5=25250
therefore average=25250/100=252.5

Like? Yes (1) | No

Rajat sadhu 10 years AGO

average=(5+10+.......100terms)/100
=>avg=((100/2)*(2*5+99*5))/100
=>avg=25250/100=252.5

Like? Yes (1) | No

Liti 10 years AGO

{100/2(5+500)}/100=252.5

Like? Yes (1) | No

sabir ahmd 10 years AGO

we know that , sum of n terms of an AP series is
=n/2(2*a+(n-1)*d) where d is common difference

here, sum=100/2*(2*5+(100-1)*5)=25250
average of this 100 multiples of 5 is=252.50

Like? Yes (1) | No

Alpana Kumari 10 years AGO

5(1+2+3+4+......+100) /100= 5*100*(100+1)/2*100=252.5

Like? Yes (1) | No

Sanju Yadav 10 years AGO

first element a=5,
last element l=500,
number of multiples n=100
s={n*(a+l)}/2*n
s={100*(5+500)}/2*100=252.5

Like? Yes (1) | No

haritha 10 years AGO

progressions---we can find last term using a+(n-1)d.from that we can find sum of n terms using AP formula n/2(a+l).we get 25250.then 25250/10=252.5

Like? Yes (1) | No

Narendra Jadon 10 years AGO

Multiples of 5 are as follows:
5 , 10 , 15 ,20 , ..................500
Sum of these 100 terms = (100/2)x( 10 + 99 x 5)
50x505 = 25250
Average = 25250/100 = 252.5
Option 7

Like? Yes (1) | No

VINAY KUMAR BHARATHA 10 years AGO

5(100*101/2)/100=252.5
because first i have found sum of first 100 multiples of 5 using (n(n+1))/2.And then i had solved

Like? Yes (1) | No

N.HAROON RASHEEDH 10 years AGO

5(1+2+...........100)/100=(5*100*101)/100*2=505/2=252.5

Like? Yes (1) | No

G.Jaya Krishna 10 years AGO

(5*((100*(100+1))/2))/100

Like? Yes (1) | No

trinadh 10 years AGO

given series in AP
Sum of the terms=n/2(a+l)
average of terms=(a+l)/2
first term=5
last term=a+(n-1)d
where a=5 n=100 and d=5
last term=500
sum=505/2=252.5

Like? Yes (1) | No

mounika kadudhuri 10 years AGO

Sum of first 100 multiples is given by A.P 5,10,....,495,500
Sum=25250
Average=sum/n
=25250/100
Average=252.5

Like? Yes (1) | No

sai 10 years AGO

5,10................500
avg = sum/ 100 =252.5

Like? Yes (1) | No

Rayapureddy V Sai Kiran 10 years AGO

average=sum/total number of terms
=(n(a+l)/2)/n
=(100(5+500)/2)/100
= 252.5

Like? Yes (1) | No

potani manikumar 10 years AGO

5(1+2+3+......100) / 100
5(5050) / 100
252.5

Like? Yes (1) | No

Preethi Kurian 10 years AGO

average of 100 mulltiples of 5=sum of 100 multiples of 5
-------------------------------------
100
=25250/100
=252.5

Like? Yes (1) | No

ananthavalli 10 years AGO

5*((100*101)/2)=25250
Average of first 100 multiples=25250/100=252.5
So option 7) is correct

Like? Yes (1) | No

pavani 10 years AGO

we know, n(n+1)/2
for 100 multiples of 5 :5(1+2+3+.........+100)/100
= 5*100*101/2*100 =505/2 =252.5

Like? Yes (1) | No

KODEISWARAN ELANGOVAN 10 years AGO

Total=100/2(2(5)+(100-1)5)
Ans=total/100

Like? Yes (1) | No

Pushpendra singh chundawat 10 years AGO

sum of 1st 100 multiple of 5 will be=5*(1+2+3.....+100)
=5*(100*(100+1))/2
so avg=5*(100*(100+1))/(2*100)
=252.50

Like? Yes (1) | No

Sneha Goel 10 years AGO

sum=5+5*2+5*3...+5*100
=5(1+2+3...+100)
=5*n(n+1)/2=25250

average= sum/100=252.5

Like? Yes (1) | No

Ankit Shrivastava 10 years AGO

first 10 no. total 5+10+15+20+25+30+35+40+45+50=275
55+60+65+70+75+80+85+90+95+100=775
and so on
(275+775+1275+1775+2275+2775+3275+3775+4275+4775)/100=252.5

Like? Yes (1) | No

Chaturbhuja Sai 10 years AGO

5*(101*100)/2*100

Like? Yes (1) | No

Arijit Acharyya 10 years AGO

sum of 100 multiples of 5
5*1+5*2+5*3+....+5*100
=5*(1+2+3+.....+100)
=5*(100*(100+1))/2
=25250
average=25250/100
=252.5
option 7)

Like? Yes (1) | No

Swapnil Shah 10 years AGO

sn=(5+500)*100/2

avg.=2525/10

Like? Yes (1) | No

Saravjeet Singh 10 years AGO

=5(1+2+3+4....+100)/100
=5*100*101/2/100
=252.5

Like? Yes (1) | No

mukesh 10 years AGO

total no=100
sum of 1,2,3,------100
sum=n(a+l)/2
so l=100
n=100
a=1
sum=100*(1+100)/2
so acording to the question avg is
avg=100*(1+100)*5/(2*100)
avg=505/2=252.5
ans 7

Like? Yes (1) | No

Ramu Giri 10 years AGO

sum of 100 no. with multiple=5*(1+2+3-----100)=25250
avg=25250/100=252.5

Like? Yes (1) | No

POULAMI CHAKRABARTY 10 years AGO

Multiples of 5 are 5,10,15,20..... upto 100th digit.this will form an A.P. series =>
5,10,15,20,25,.......,500 with c.d 5.
We know,
sum of an A.P. series,S =(n/2)*[(2*a)+(n-1)*d]
where, n=no. of terms
a=first term
d=c.d.
applying the formula we get S =(100/2)*[(2*5)+(100-1)*5]
=50*505
hence their average will be=>S/100
=50*505/100
=252.5
hence option c is correct.

Like? Yes (1) | No

mithun mondal 10 years AGO

the number will be 5+10+15+20+...........+100 terms
formula n/2(2a+(n-1)d)
100/2(2*5+(100-1)*5)=252.5

Like? Yes (1) | No

gouthamdaneti 10 years AGO

the sum of first 100 multiples of 5 is 5*(100*101)/2
average=sum/100

Like? Yes (1) | No

Rahul shaw 10 years AGO

5*100*101/200=252.5

Like? Yes (1) | No

fouzia 10 years AGO

5+10+15+...............+500
=5(1+2+3+4............100)

sum of N terms= n/2[2a + (n-1)d]
n=100
d=2-1=1
a=1

sum=100/2[2*1 + (100-1)1]
=50[2+99*1]
=50*101
=5050
Now 5* Sum
=5*5050
=25250

averAGE OF 100 NUM means divided by 100

25250/100= 252.50=252.5
ANS=252.5

Like? Yes (1) | No

Dikesh Kumar Gumansingh 10 years AGO

Addition of 5 to 500 is 25250
so avg is 252.5

Like? Yes (1) | No

T.Aravinth kumar 10 years AGO

5+10+......500=25250
25250/100=252.5.

Like? Yes (1) | No

SAYAN CHATTERJEE 10 years AGO

5+10+15+.......................100th terms
s=100/2(2*5+(100-1)*5)
so, average=s/100= 252.5 (7)

Like? Yes (1) | No

J.Rupesh kumar 10 years AGO

5(1+2+3+4+.........100)/100

5(100*101)/(2*100)

505/2=252.5

Like? Yes (1) | No

suman kumar 10 years AGO

avg of 100 multiple=sum of no's/no. of numbers
=>avg=(5+10+15+20+_ _ _500)/100
=>avg=5*(1+2+3+4+ _ _ _)/100
=>avg=(5*100*101)/(2*100)
=>avg=252.5
so, option 7 is correct.

Like? Yes (1) | No

NIRUPOM GHOSH 10 years AGO

5x1 + 5x2 + 5x3 + 5x4 ..... 5x100 = 5 * (1 + 2 + 3 + 5 .... + 100)= 5 * (sum of n numbers).
For calculating the sum of n numbers we have the formula i.e (n * (n + 1)) / 2 apply the formula in the above equation.where n = 100 so= 5 * (100 * (100 + 1)) / 2= 5 * (50 * 101)= 5 * 50 * 101= 25250 This the sum of the first 100 positive number which is multiples of 5.And for average:= 25250 /100= 252.50

Like? Yes (1) | No

DIVANSHU 10 years AGO

sn=(5+500)*n/2
=25250/100
=252.5

Like? Yes (1) | No

kantheti.navya 10 years AGO

5+10+............................+5(100)
5(1+2+............................+100)
average:5(100*101/2)/100
=505/2
252.5

Like? Yes (1) | No

Naresh 10 years AGO

5 10 15 . . .100 terms
sum of n terms in AP=n/2(2a+(n-1)d)
where a=5;n=100;d=5(common diff)

Like? Yes (1) | No

Siva Prakash 10 years AGO

first 100 multiples of 5

i.e., 5*1 + 5*2 + 5*3 +........+ 5*100

=> 5* (1+2+3+...+100)

=> 5* [(100 (100+1)) / 2]

=> 5 * 5050

= 25250

Now calculate average

i.e., 25250/100 = 252.5

Like? Yes (1) | No

makthabai akash 10 years AGO

it is for average of first 100 multiples of 5
so avg for 1st multiple will be 5/1=5
and for 2multiples of 5 it will be (5+10)/2=7.5
and for 3 multiples (5+10+15)/3=10
so the series will be like 5,7.5,10,12.5,15.....
that is for every increase in multiple 2.5 is added for the previous avg.
if the number of multiples to average is n,then
nth average is written as 5+[(n-1)2.5]
so for 100 multiples it is 5+99(2.5)=252.5

Like? Yes (1) | No

adarsh kumar yadav 10 years AGO

5(1+........100)=
5*100*101/2
now its 25250
now take average of 100 no so divide it by 100
252.5

Like? Yes (1) | No

vishnuvardhan 10 years AGO

100/2*(5+500)*(1/100)=252.5

Like? Yes (1) | No

p.jaya satish chandra 10 years AGO

(100/2)*(5+500)/2*(1/100)=252.5

Like? Yes (1) | No

veeranjaneyulu 10 years AGO

because average of 5 and 500=252.5 likewise 10 and 495=252.5

Like? Yes (1) | No

shiv pujan 10 years AGO

sum of 100 multiple of 5
=5*(1+2+3......+100)=5*5050
avg =5*5050/100=252.5
hence option 7.

Like? Yes (1) | No

balagopalakrishna 10 years AGO

5,10,......500

(100/2)(505)(1/100)=252.5

Like? Yes (1) | No

shahil raj 10 years AGO

THE SERIES IS FORM AS

5,10,15,.... SO ON UP TO 100 TERM
THE SERIES IS IN A.P WITH C.D OF 5
SO,
Sn= N/2{A+(N-1)D}
Sn=99/2{5+99*5}=252.5

Like? Yes (1) | No

PRIYANK BHARGAVA 10 years AGO

50(5+500) == sum of terms
dividing by 100 we get
252.5 .............ans

Like? Yes (1) | No

manisha 10 years AGO

no's would be :- 5, 10,15,20 .................. ,500
this will be in arithmetic progression series, with common difference(d) 5 ;
now, we know.. 1st term is 5 i.e a=5, d=5, (last term) l = 500;
=> sum = n/2 (a+l) ;
=> s= 100/2(5+500);
=> s=50*505
=> s = 25250
now average will be;
sum of all no's/ total no,
=> 25250/100 ;
average = 252.50 (ans)

Like? Yes (1) | No

TUSHAR 10 years AGO

252.5

Like? Yes (1) | No

manasa 10 years AGO

(average is always middle number of the given integers)
in case of odd it is exactly middle num while in case ofeven it is average of sum of the two middle numbers ex: in case of 100 (even) multiples of 5 5*50=250,58*51=255 their sum =505 avg=252.5

Like? Yes (1) | No

nisha kumari singh 10 years AGO

sum of 1st 100 multipels of 5 is =100/2(2*5+99*5)=25250
average=25250/100=252.5

Like? Yes (1) | No

vinay kumar varanganti 10 years AGO

((1/1-5)/99)*1000

Like? Yes (1) | No

Saurabh Mehta 10 years AGO

Its an AP,find the sum using AP formula and then divide it by 100.

Like? Yes (1) | No

sumit mondal 10 years AGO

5*n(n+1)/2

5*100(100+1)/2

5*50*101

now we need to find average so.

(50*5*101)/100=252.5

Like? Yes (1) | No

dushyant singh pawar 10 years AGO

sum=100/2(10+(100-1)*5)

sum=25250

average=sum/100=252.5{Answer 7}

Like? Yes (1) | No

uday kumar reddy 10 years AGO

sumofterms in A.P/total no of terms

Like? Yes (1) | No

DIPAK TIWARI 10 years AGO

Avg of first 100 multiples of 5:
=5(1+2+3+.....+100)/100
=(5*100*101)/(2*100)
=505/2
=252.5
Option 7

Like? Yes (2) | No (2)

Ravi Bhalani 10 years AGO

5+10+15+20+....500=25250
Average=25250/100=252.5

Like? Yes (1) | No (1)

vaishnavi 10 years AGO

first 100 multiple of 5 =5,10,15,.....500
average=(5+500)/2=505/2=252.5

Like? Yes (1) | No (1)

Britto prabhu 10 years AGO

5*(1+2+3+.........+100)

sum of n numbers=n(n+1)/2

n=100

5*(100*101)/2=250*101=25250

avg of 100 numbers = 25250/100

252.50

Like? Yes (1) | No (2)

Dimple

India

Punjab

Time: 00:01:33

Points

Maths Quotes

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Plato

Mathematics is the brain's sharpener.

Angeliza Ampeloquio, PH

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What is the average of first 100 multiples of 5.

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