Find the sum of all natural number between 250 and 1000 which are divisible by 3.

Vaibhav Mishra   11 years AGO

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999
This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999
We Know that
l=a+(n-1)d
999=252+(n-1)3
999 - 252 = 3(n - 1)
n=250
We know that
S=n/2(a+l)
=250/2(252+999)
=125*1251
=156375
Hence, the required sum is 156375.(Ans)

Like? Yes (43) | No (9)  

nishant kumar   11 years AGO

252,255, - - - - - - -999
a=252
d=3
an=999
an= a+(n-1)d
999=252+(n-1)3
999=252+3n-3
3n=750
n=250
=>sn=n/2(a+an)
sn=250/2(252+999)
=>sn=156375

Like? Yes (15) | No (4)  

Madhu pericherla   11 years AGO

252,255.....999
a=252
d=3
an=999
an=a+(n-1)d
an=252+(n-1)3
999=252+(n-1)3
n=250
sum of all natural numbers
sn=n/2(a+an)
sn=250/2(252+999)=156,375
sum of all natural numbers is 156,375

Like? Yes (14) | No (6)  

SOMASHEKAR MACHANI   11 years AGO

numbers are 252,255,......999
it is in ap
999 = 252+(n-1)3
n=250
sum=(250/2)*(252+999)
sum=156375

Like? Yes (11) | No (4)  

praveen sb   11 years AGO

by arithmetic progression we take a=252
l=999 because it is last term ie divisible by 3 & d=3
t(n)=a+(n-1)d
999=252+(n-1)3, we get n=250
hence by formula s(n)=n/2(a+l)=125(252+999)=156375

Like? Yes (7) | No (1)  

Mariappan   11 years AGO

The required sum is
252+255+....+999

Therefore, a1=252, an=999

so, d=3

To Find n,

WKT, an=a1+(n-1)d

=> n=250

WKT, Sn=(n/2)*(a1+an)

Therefore, S250=(250/2)*(252+999)

=> S250= 156375 (Ans)

Like? Yes (6) | No   

Aniruddha Bhattacharjee   11 years AGO

a=252 (as it is next number divisible by 9)
d=3
l=999 (as last number divisible by 9)
to find n
999=252+(n-1)3
747=3n-3
750=3n
n=250
S250=250/2(252+999)=156375

Like? Yes (8) | No (3)  

Rashmi bharti   11 years AGO

a1= 252,
last number =999=an
999=252+(n-1)3,
747/3=(n-1)
250=n
now sum of series = n/2(a1+an)
S= 125(252+999)=156375

Like? Yes (6) | No (2)  

KUNAL GAURAV   11 years AGO

Between 250 and 1000 we have first term divisible by 3 =252 and last term = 999

So total no. of terms = ((999-252)/3) +1 = 250

Putting n=250 , a=252, l=999 in the formula
sum = n/2[a+l]

we get sum = 156375

Like? Yes (5) | No (1)  

Raju Sharma   11 years AGO

tn=a+(n-1)*d
999=252+(n-1)*3 then n=250
Sn=n/2(a1+an)
then solve ans =156375

Like? Yes (5) | No (1)  

praneeth   11 years AGO

252,255,258,261,.......999.
the series is in Arithmetic progression with a=252;l=999;d=3.first we shall find n using formula l=a+(n-1)d.

999=252+(n-1)3;n=250;
therefore,sum=n/2(2*a+(n-1)d)=125(2*252+249*3);
125(504+747)=1251*125=156375

Like? Yes (4) | No (1)  

AISHWARYA   11 years AGO

The natural numbers between 250 to 1000 which are divisible by 3 are 252,255,......,999
a1=252
d=3
a1+(n-1)d substitute the values will get n=250
sn=n/2(a1+an)
s250=250/2(252+999)
hence the required sum=156375

Like? Yes (3) | No   

Sudharsan   11 years AGO

clearly the sequence is AP with a=252 and d=3 and last term is 999
number of terms = ((999-252)/3)+1=250
sum of the terms is n/2(2a+(n-1)d)=156375

Like? Yes (4) | No (1)  

ANGELIN ALFRED   11 years AGO

sum of natural no.s div by 3,S=(sum of no.s div by . 3 from 1to1000,s1)
-(sum of no.s div . by 3 from 1to . . 254)
s1=3(1+2+...+333)=166833
s2=3(1+2+...+83)=10458
s=s1-s2=156375

[sum of nat no. =(((n*(n+1))/2)
where n is last no. ]
here n1=333 and n2=333

Like? Yes (3) | No   

Abhishek misra   11 years AGO

series are 252,255,258,........999
a=252,d=3,l=999
l=a+(n-1)*d
999=252+(n-1)3
999=252+3n-3
n=250
sum=n/2*(a+l)
sum=250/2(252+999)
sum=125*1251
sum=156375

Like? Yes (4) | No (1)  

Dineshkumar M   11 years AGO

The first term which ll divide by 3 is 252 and the last term which ll divide by 3 is 999.

a=252 {Which is the first term in AP}

The following formula is used to find nth term in AP series.
an=a+(n-1)d

Here d=3 and nth term is an=999

By using the above formula we can find value for n=250.

By using theformula for sum of n number is
Sn= n/2[2a+(n-1)d]

sn=156375

Like? Yes (4) | No (1)  

KALIDASS   11 years AGO

let the series be
=(3+6+9+...+999)-(3+6+9+....+249)
=3(1+2+3+.....+333)-3(1+2+3+....+83)
=3*(333*334/2)-3*(83*84/2)
=166833-10458
=156375.

Like? Yes (4) | No (1)  

Udhaya kumar   11 years AGO

from 1..........250=3(83)(84)/2=10458
from 1...........1000=3(333)(334)/2=166833
166833-10458=156375.....

Like? Yes (4) | No (1)  

FAISAL RIZWAN   11 years AGO

n*(n+1)/2
n=1000
so on solving we get 500500 so for getting actual value we have to subtract the sum of 1 to 250
now n=250 so apply n*(n+1)/2
then the value will 31375
Now subtract 500500-31375=469125

So for dividing 469125/3 =156375 will be exactly divisible by 3

Like? Yes (4) | No (1)  

Mayuresh Srivastava   11 years AGO

Between 250 to 1000 nos. divisible by 3 are
252,255,258,.......999.
This is an A.P. with a=252 d=3
from 999=252+(n-1)3 we have n=250
now sum=250/2(252+999)=156375.

Like? Yes (4) | No (2)  

sanjay kumar saha   11 years AGO

fist and last term between 250 and 1000 which is divisible by 3 is 252 and 999.
999=252+(n-1)*3
n=250
sum=250/2(2*252+249*3)=156375

Like? Yes (3) | No (1)  

richa ahuja   11 years AGO

252,255....999
it will form a AP
with first term 252 , last term 999 and common difference 3
l=a+(n-1)d
999=252+(n-1)*3
n=250
sn=n/2(2a+(n-1)d)
putting values of n,a,d
sn=156375

Like? Yes (3) | No (1)  

ch.jyothsna   11 years AGO

series is 252,255,........,999 series is in A.P a=252 d=3 l=999 to find n we should solve 999 = 252+(n -1)3 then we will get n= 250
now substitute all a,n,l,d in formula n(a+l)/2 = 250(252+999)/2 = 156375

Like? Yes (4) | No (2)  

MAN   11 years AGO

all natural number between 250 & 1000 which are divisible by 3 is,
252,255,258,261,264..............999
since these are in A.P with first term=252 and d=3,
nth term,Tn=252+(n-1)*3
999=252+(n-1)*3
on solving,we get
n=250;
now,Sn=n/2*(a+Tn)
Sn=250/2*(252+999)
on solving,we get
Sn=156375 ans.

Like? Yes (3) | No (1)  

Chandni Maurya   11 years AGO

ans=(9).156375
the 1st and last no between 250 and 1000 which is divisible by 3 is= 252,999 resp.
consider AP series, a=252, Tn=999, d=3
Tn=a+(n-1)d
999=252+(n-1)3
=>n=250
Sum=n/2(2a+(n-1)d)
=250/2(2*252+(250-1)3)
=156375

Like? Yes (3) | No (1)  

MUKESH KUMAR   11 years AGO

n=(999-252)/3 +1=250
sum=250/2(2*252+249*3)

sum=156375

Like? Yes (3) | No (1)  

Jacky shevkani   11 years AGO

first we should know the value of n in the ap
therefore
tn=a+(n-1)d
we know tn is 999 and d is 3
substituting in above equation we get
n=250
now to find sum of all the terms between 250 and 1000 which are divisible by 3 we should use the following formula
sn=(n/2)/(t1+tn)
substituting we get
sn=156375

Like? Yes (4) | No (2)  

Aditya C S   11 years AGO

The first number is 252 and last number is 999.
Tn = a + (n-1)*d
Tn = 999
a = 252
d = 3
So solving the above equation n = 250
Sum of the numbers = n*(a+l)/2
n = 250
a = 252
l = 999
sum = 156375

Like? Yes (3) | No (1)  

ravi singh   11 years AGO

answer is 156375

here first term and last term which is divisible by 3 is 252 and 999

by using formula of AP
a+(n-1)*d =L

where L(last term) =999
a(first term) =252

and d is difference =3
n is no. of terms

hence, 252 +(n-1)*3 =999

n= 250

hence sum= n*(a+L)/2=156375

hence answer is option 9

Like? Yes (3) | No (1)  

Balakrishnan M   11 years AGO

sum of all natural number between 250 and 1000 which are divisible by 3,
252+255+258+.....+999
Number of terms=n=[(l-a)/2]+1=[(999-252)/2]+1
=250
sum=(n/2)[l+a]
=(250/2)[999+252}
=(125)(1251)
=156375

Like? Yes (3) | No (1)  

diwakar kumar singh   11 years AGO

between 250 and 1000 the first no divisible by 3 is 252 and the last no is 999.so
s=(3*84+3*85+3*86..............+3*333)
=3(84+85+86...................+333)
=3(250*84+125*249)
=156375

Like? Yes (5) | No (3)  

Suman Saurabh   11 years AGO

first term = 252 & last term = 999

so,999=252+(n-1)3
n=250

sum=n/2 * (first term +last term)

=250/2 * (252+999)
=156375

Like? Yes (3) | No (1)  

sanjay singh   11 years AGO

the first and last numbers in the series, divisible by 3 are 252 and 999
by a+(n-1)*d=999 we get n=250
by n(2a+(n-1)d)/2 and putting n=250
we get n=156375

Like? Yes (3) | No (1)  

Yallapu Uday tarun kumar   11 years AGO

the no's that are divisible by 3 b/w 250 and 1000 are 252,255,258,.......,999=3(84,87,....,999)
sum=3((333*334)/2-(83*84)/2)=3(52125)=156375.

Like? Yes (3) | No (1)  

DIPAK KUMAR BHAGAT   11 years AGO

the 1st integer, after 250 which is divisible by 3 is 252 & the last integer before 1000 which is divisible by 3 is 999.
therefore, the sequence of integers b/w 250 & 1000 which are divisible by 3 is,
252,255,258,.......,999.
now, it is an a.p. with
a=252, d=3
so, nth=a+(n-1)d
999=252+(n-1)*3
=> n=250;
now, sum of n terms is,
sum=n(a+l)/2 (here l is last integer=999)
sum=125*1251=156375;

Like? Yes (3) | No (1)  

LOGAN WOLVERINE   11 years AGO

the series is 252+255+.......+999;
sum=n/2(a1+an);
no of terms is 751/3=250(approximate);
sum=250/2(252+999);
=125(1251);
=156375.
option 9.

Like? Yes (2) | No   

kishore kumar   11 years AGO

series: AP
252,255,258,...,999 (252 is the first number that is divisible by 3 so its first term)
now,
a=252, d=3, tn=999;
from,

tn=a+(n-1)*d
999=252+(n-1)3
solving we get,
n=250..
we know nth term and n.. so
Sn= (n/2)(a+an)
=(250/2)(252+999)
=156375

Like? Yes (2) | No   

ponprakash   11 years AGO

ans 9)156375
252,255,..............999
a=252
d=3
an=999
an=a+ (n-1) d
999=252+(n-1)3
n=250
s=n/2(a+an)
s=125(252+999)
s=156375..

Like? Yes (2) | No   

LAKSHMI   11 years AGO

250,....,1000
252,255,258,.....,999
we have t=a+(n-1)d
a=252 d=3 t=999

999=252+(n-1)3

n=250

sum=(n/2)(1st term + last term)
sum=(125)(252+999)
sum =156375

Like? Yes (2) | No   

shivendra pratap singh   11 years AGO

252,255,...............999
t=a+(n-1)d
999=252+(n-1)3
999=252+3*n-3
999=249+3*n
999-249=3*n
750=3*n
n=250
s=250/2[2*252+(250-1)3]
125[504+249*3]
=156375
so option 9) 156375
is correct

Like? Yes (3) | No (1)  

sainath   11 years AGO

Sum of and AP is n/2(a+l) where
a - first term
l - last term
n - number of terms
Here a=252 and l=999
and n = (999-252)/3 + 1 = 747/3 + 1 =249+1 = 250
[Since we have to include last term also we added 1]
So sum is 250/2 * (252+999)
=125*1251 = 156375

Like? Yes (4) | No (3)  

abhishek ranjan   11 years AGO

1st no. exactly divisible by 3 is 252 and lat term exactly divisible by 3 is 999.

first term, a = 252

last term, Tn= 999

common difference = 3

applying Arithmetic progression formula-,

Tn = a + (n-1)d
999 = 252 + ((n-1)3)
=> n = 250

sum of n terms of AP = (n(a+Tn))/2

sum = (250(252+999))/2

= 156375

Like? Yes (2) | No (1)  

DILIP KUMAR   11 years AGO

Numbers divisible by 3 b/w 250 and 1000 are : 252,255....999
first num = 252
last num =999
d=3
its in AP
252+(n-1)3=999
solving we get n= 250
sum, Sn=(n/2)(first Num + Last Num)
Sn=(250/2)(252+999)
=125*1251
=156375

Like? Yes (3) | No (2)  

Deepika dabgotra   11 years AGO

a.p::252,255,258...........999
l=999
a+(n-1)d=999
252+(n-1)d=999
from this equation we get n=250
now by using the formula sn=n/2(a+l)
we get sum=250/2(252+999)
=125*1251=156375

Like? Yes (2) | No (1)  

Parthiban C   11 years AGO

n=a+(n-1)d
s=n/2(first+last term)
s=250/2(252+999)
=156375

Like? Yes (3) | No (2)  

Balakumaran N.A   11 years AGO

AP : 252 , 261 , - - - - - - - - - - - - - - - - - - - - - - 999
a = 252
d = 3
an = 999
an = a + (n-1) d
999 = 252 + (n-1) 3
999 = 252 + 3n - 3
999 = 249 +3n
750 = 3n
250 = n


Sn = n/2 (a +an)
Sn = 250/2 (252 +999)
Sn = 125 (1251)
Sn = 156375

Like? Yes (2) | No (1)  

ShAg   11 years AGO

Natural numbers between 250-1000 divisible by 3 are:
252,255,258....999 which are in A.P.
common difference 'd' = 3
number of terms n = (1000-250)/3 = 250
so, sum S=(250/2)(252+999) = 156375

Like? Yes (2) | No (1)  

mandira kundu   11 years AGO

total no. of numbers btn 250 to 1000 are 250.
so n/2(a1+an) where n=250,a1=252,a=999 we get sum=156375

Like? Yes (4) | No (3)  

Harleen Chadha   11 years AGO

series 252, 255, .........., 999
Tn = a + (n-1)d
999 = 252 + (n-1)*3
n = 250
sum = [n * (a + Tn)]/2 = 156375

Like? Yes (2) | No (1)  

k.d.n padma rao   11 years AGO

natural numbers between 250 and 1000 which are divisible by 3 are:252,255,258,...999
here a=252,d=3
sum=(na+(n(n-1))/2)d=156375

Like? Yes (2) | No (1)  

komal   11 years AGO

a=252;l=999,n=250
s=n/2(a+l)=156375

Like? Yes (2) | No (1)  

Churamani Prasad   11 years AGO

total nos= 1000-250 = 750
now,750\\3 = 250=n (let)
let a=252 & l=999

sum= n\\2 * (a+l)
250\\2*252+999= 156375

Like? Yes (2) | No (1)  

Mayank shyamsukha   11 years AGO

252=1st no
999=last no
999=252+(n-1)3
solving n=250
s=n/2(a+l)
or,s=125(252+999)=156375

Like? Yes (2) | No (1)  

saurabh saraff   11 years AGO

nth term = 999 and first term is 252 as divisible by 3 is said.
so 999 = 252 + (n-1)* 3
or n = 250..

Sum of n terms = n/2 ( a + l)
= 250/2 * (252 + 999 ) = 125 * 1251 = 156375

Like? Yes (2) | No (1)  

bindu   11 years AGO

252,255,...999 is the series divisible by 3
so find out number of these using arthemic progession formula
tn=a+(n-1)d;
999=252+(n-1)3
we get n=250
now find sum using formula sn=n/2(2a+(n-1)d)
sn=250/2(2(252)+249*3)
we get sn=156375

Like? Yes (2) | No (1)  

Devendra Marghade   11 years AGO

number between 250 and 1000 divisible by 3 are 252,255,258,......,999
To find number of terms, 999=252+(n-1)3, n=250
Sum of 250 terms with a=252 & d=3
=(250/2)(2*252+249*3)=125(1251)=156375

Like? Yes (2) | No (1)  

ashiwani kumar singh   11 years AGO

a=252 l=999 tn=999 d=3
tn=a+(n-1)*d
n=250;
sn=n/2*(a+l)
sn=156375

Like? Yes (2) | No (1)  

shailesh kumar mishra   11 years AGO

the no bet 250 to 1000 is 750(1000-250)
then the no divisibal by 3 is 750/3=250
so n=250
and 252 is the fist no of series so a=252
now since all no multipal of 3 so diff d=3
now put all in S= n/2( 2a+(n-1)d )=156375 is the ans

Like? Yes (2) | No (1)  

mohamed ismail   11 years AGO

a1=252 ,an=999,d=3
sum =[n/2] *(a1+an)

Like? Yes (2) | No (1)  

Sagar Jinachandran   11 years AGO

AP : 252 , 261 , - - - - - - - - - - - - - - - - - - - - - - 999
a = 252
d = 3
an = 999
an = a + (n-1) d
999 = 252 + (n-1) 3
999 = 252 + 3n - 3
999 = 249 +3n
999-249 = 3n
750 = 3n
756/3 = n
n = 250

Sn = n/2 (a +an)
Sn = 250/2 (252 +999)
Sn = 125 (1251)
Sn = 156375
option 9

Like? Yes (3) | No (3)  

Muhammad Nadeem Haider Naseem   11 years AGO

Sn=n(l+a)/2
where Sn is the sum of the terms, n is the number of terms, l is the last term, and a is the first term.

a= 252 (the first number after 250 divisible by 3)
l= 999 (the last number before 1000 divisible by 3)

252/3 = 84
999=3 = 333

therefore, n= 333-84+1 = 250

So,
Sn=250(999+252)/2
=250*1251/2
=312750/2

Like? Yes (2) | No (2)  

IFFAT IRAM   11 years AGO

first term which is divisible by 3 = 252
last term=999
common difference=3
999=252+(n-1)*3 where n=no. of terms
747/3=n-1
249+1=n
n=250
sum=250/2[252+999]
125*1251
156375

Like? Yes (2) | No (2)  

SHIVANI PRAKASH   11 years AGO

no.of terms=250
sum=250/2[252+999]=156375

Like? Yes (2) | No (2)  

sudipta sadhukhan   11 years AGO

the first term will be 252 & last term will be 999
& their are 250 term which are divisible by 3.
so their summation 250/2*(252+999)=156375

Like? Yes (1) | No (1)  

Gokul Krishnan   11 years AGO

l=a+(n-1)d
n=250
s=(n/2)(a+l)
s=156375

Like? Yes (1) | No (1)  

saima kidwai   11 years AGO

first no. & last no. btwn 250 & 1000 are 252 & 999 resp which are dvsble by 3,
999=252+(n-1)3 ; l=a+(n-1)d
n=250 is the no. of numbers dvsbe by 3
sum,S=n/2(2a+(n-1)d)
S=250/2(2*252+(250-1)*3)
S=156375

Like? Yes (2) | No (2)  

saravanakumar   11 years AGO

A=252 D3 L 999
a+n-1d 999
252+n-13 = 999
n-1 3 =747
n=250
sum=156375

Like? Yes (1) | No (1)  

GOKULVASANTH   11 years AGO

3((1+2+....+333)-(1+2+...+83))=156375

Like? Yes (2) | No (2)  

Thejashwini G   11 years AGO

this is arithmetic progration ,first we have to find out numbers between 250 and 1000 which are divisible by 3,i.e., between 252 and 999 by using formula Tn=a+(n-1)d.After finding n substitute it in the formula Sn=n/2(a+l)where,a is the first term,l is the last term and d is the difference between the term and its previous term

Like? Yes (1) | No (1)  

arjun   11 years AGO

ans 156375
252, 257, 263, 269, 271, 277, 281, 283,
293, 307, 311, 313, 317, 331, 337, 347,
349, 353, 359, 367, 373, 379, 383, 389,
397, 401, 409, 419, 421, 431, 433, 439,
443, 449, 457, 461, 463, 467, 479, 487,
491, 499, 503, 509, 521, 523, 541, 547,
557, 563, 569, 571, 577, 587, 593, 599,
601, 607, 613, 617, 619, 631, 641, 643,
647, 653, 659, 661, 673, 677, 683, 691,
701, 709, 719, 727, 733, 739, 743, 751,
757, 761, 769, 773, 787, 797, 809, 811,
821, 823, 827, 829, 839, 853, 857, 859,
863, 877, 881, 883, 887, 907, 911, 919,
929, 937, 941, 947, 953, 967, 971, 977,
983, 991, 997
total numbers 115

so a.p=a+(n-1)d

sub and get this ans

Like? Yes (1) | No (1)  

Abhishek   11 years AGO

252+(n-1)*3=999
n=250
250/2*(252+999)
=156375

Like? Yes (3) | No (4)  

manasa epuru   11 years AGO

252+255+..............+999
3(84+85+............+333)
3((1+2+.....+333)-(1+2+.....+83))
3[(333(334)/2)-83(84)/2)]
=156375

Like? Yes (2) | No (3)  

K SUMAN RAO   11 years AGO

first term=252
last term=999
no. of terms=(1000-250)/3=250
sum=n/2(2a+(n-1)d)=125*(504+249*3)=156375

Like? Yes (1) | No (2)  

Tanu   11 years AGO

First term=252
Last term=999
Sum=156375

Like? Yes (1) | No (2)  

chandan kumar yadav   11 years AGO

here ist term, a=252
last term , b=999
no. of terms between 1000 and 250 which are divided by 3 are 250..
sum of the ap=250/2(2*252+(250-1)3)
Hence the sum is 156375.

Like? Yes (1) | No (3)  

Shubham Kumar   11 years AGO

solution=(sum of all no from 1 to 1000,divisible by 3)-(sum of all no from 1 to 250,divisible by 3)
i.e equal to (3+6+9+.....999)-(3+6+.......249)

Like? Yes (1) | No (3)  

Shuvankar Ghosh   11 years AGO

using formula of AP series

Like? Yes (1) | No (3)  

jaffreyhudson   11 years AGO

simply solve the sum

Like? Yes (1) | No (3)  

Hemant Kumar   11 years AGO

252
255
258
261
264
267
270
273
276
279
282
285
288
291
294
297
300
303
306
309
312
315
318
321
324
327
330
333
336
339
342
345
348
351
354
357
360
363
366
369
372
375
378
381
384
387
390
393
396
399
402
405
408
411
414
417
420
423
426
429
432
435
438
441
444
447
450
453
456
459
462
465
468
471
474
477
480
483
486
489
492
495
498
501
504
507
510
513
516
519
522
525
528
531
534
537
540
543
546
549
552
555
558
561
564
567
570
573
576
579
582
585
588
591
594
597
600
603
606
609
612
615
618
621
624
627
630
633
636
639
642
645
648
651
654
657
660
663
666
669
672
675
678
681
684
687
690
693
696
699
702
705
708
711
714
717
720
723
726
729
732
735
738
741
744
747
750
753
756
759
762
765
768
771
774
777
780
783
786
789
792
795
798
801
804
807
810
813
816
819
822
825
828
831
834
837
840
843
846
849
852
855
858
861
864
867
870
873
876
879
882
885
888
891
894
897
900
903
906
909
912
915
918
921
924
927
930
933
936
939
942
945
948
951
954
957
960
963
966
969
972
975
978
981
984
987
990
993
996
999
total = 156375

Like? Yes (4) | No (8)