(1+3+5+7+...........+3983) / 1992 is equal to

Richa Saini   8 years AGO

Equation: (1+3+5+7+......+3983)/1992
As 1 3 5 7..... 3983 are in AP series.
Therefore ,Sum of AP series = (first term +last term) * number of terms/2

Number of terms= no of odd numbers from 1 to 3983
ie: 1992 (half of total numbers)
=>Sum=(1+3983)*1992/2
=>1992*1992

Then puting this in above equation-
(1992*1992)/1992
=> 1992
Therefore, option(6) is the answer.

Like? Yes (8) | No   

MUKESH KUMAR   8 years AGO

1+3+5.....3983 is in A.P with d value=2
so a+(n-1)d=3983
(n-1)2=3983-1
n-1=1991
n=1992
we know that sum of first n odd digits= n^2
so 1992*1992/1992=1992

Like? Yes (4) | No   

Yashi Das   8 years AGO

Number of odd no.s from 1 to 3983 is 1992
therefore,((1992/2)(1+3983))/1992
=3984/2 =1992

Like? Yes (3) | No   

satish ojha   8 years AGO

1,3,5.........3983.. is ap .
so no of terms
Sn=a+(n-1)d

3983=1+(n-1)2
3984=2n
n=1992
sum =1992(1+3983)/1992*2= 3984/2=1992

Like? Yes (2) | No   

RAKESH   8 years AGO

n = (3983+1)/2 = 1992
(1+3+5+7+...........+3983) / 1992
= 1992^2 / 1992
= 1992

Like? Yes (1) | No   

MRITUNJAY   8 years AGO

(1+3+5+...........+3983)/1992
from series
tn=a+(n-1)d
3983=1+(n-1)*2
after solving we get n=1992
then sum=1992/2*(1+3983)
=1992^2
from question
1992^2/1992=1992
option 6

Like? Yes | No   

RAHUL KUMAR   8 years AGO

Given series in an A.P in
which a=1,d=(3-1)=2,l=3983
let the no of terms in it be
n.then
nth term=a+(n-1)d
1+(n-1)2=3983
n=1992
required sum=n/2(a
+l)=1992/2(1+3983)=1992x1992
ans will be
1992x1992/1992=1992
option (6) is r8

Like? Yes | No   

Mukund Manohar   8 years AGO

(1+3+5+7+.....3982) is in AP with a=1,d=2 and l=3983
Now, number of terms,n=(1+3983)/2=1992
Sum of AP=(n/2)*(a+l)=(1992/2)*(1+3983)=(1992/2)*3984
Answer=(1992*3984)/(2*1992)=1992

Like? Yes | No   

Devendra Marghade   8 years AGO

If 'n' are the number of terms in the series, then
1+(n-1)*2=3983, n=1992
So sum of these terms/1992=(1992/2)[2*1+(1992-1)*2]/1992 =(1/2)(2+3982)=1992
Option 6)

Like? Yes | No   

ABHINAV PUSHPAM   8 years AGO

Let the number of terms in 1 + 3 + 5 + ... + 3983 be n
Then 1 + (n ? 1) × 2 = 3983 ? n = 1992
? 1 + 3 + 5 + ... + 3983 =
1992
2
× (1+ 3983) = 1992 x 1992.
?
1+3+5+7... +3983/1992= 1992*1992/1992
=1992 ...
option 6)

Like? Yes | No   

Saurabh Bajaj   8 years AGO

No. of terms = 3983/2 = 1991.5 = 1992
Sum of first n odd terms = n2
In this case, 1992*1992/1992 = 1992

Like? Yes | No   

Amit   8 years AGO

Number of terms=1992
Sum=1992/2(2*1+1991*2)=1992*1992
So ans=1992

Like? Yes | No   

Bablu patel   8 years AGO

2n=3983+2-1
n=1992
sum=(1+3983)*1992/2*1992
1992

Like? Yes | No   

mcamanjeet   8 years AGO

common.difference=2
sum=n/2*(first term+last term)=n/2*(a+l)
=1992*(1+3983)/2
A/c
=(1992*3984)/1992*2
=1992

Like? Yes | No   

Sagar Jinachandran   8 years AGO

Sum of odd numbers till 3983 =3968064
3968064/1992=1992
Option 6

Like? Yes | No   

sugayathri   8 years AGO

1+3=4, 2^2
1+3+5=9, 3^2

1+3+5+7=16, 4^2
1+3+.............+3983

a=1
d=2
l=3983
l=a+(n-1)d
3983=2n-1
n=1992
n^2
1992^2

=1992^2/1992
=1992

Like? Yes | No   

MAMTA KUMARI   8 years AGO

SUM=n*(A+L)/2*1992
=(1+3983)
=3984/2
=1992
n=1992
sum=1992

Like? Yes | No   

Dheeraj kumar prajapati   8 years AGO

tn=a1+(n-1)d => n = 1992
s = n/2(2a + (n-1)d) = 1992

Like? Yes | No   

Maneesh Gupta   8 years AGO

Given series in an A.P in which a=1,d=(3-1)=2,l=3983
let the no of terms in it be n.then
nth term=a+(n-1)d
1+(n-1)2=3983
n=1992
required sum=n/2(a+l)=1992/2(1+3983)=1992x1992
ans will be 1992x1992/1992=1992

Like? Yes | No   

Shagun Chaudhary   8 years AGO

Given series in an A.P in which a=1,d=(3-1)=2,l=3983
let the no of terms in it be n.then
nth term=a+(n-1)d
1+(n-1)2=3983
n=1992
required sum=n/2(a+l)=1992/2(1+3983)=1992x1992
ans will be 1992x1992/1992=1992

Like? Yes | No   

ravi   8 years AGO

sum of n odd numbers is n^2. so find the number of odd numbers. using A.P 3983=1+(n-1)2
==> n=1992
==> n^2/1992=1992.

Like? Yes | No   

Akshay Nivrutti Raskar   8 years AGO

Addition of n consecutive odd numbers = n * n
Addition of (3984/2 = 1992) odd numbers = 1992 * 1992

Required ans = 1992 * 1992 /1992 = 1992

Like? Yes | No   

YUGANDHARA RAO    8 years AGO

By using the formula for Sum of "n" odd numbers as n^2.
We will get n=1992 as n=(3983+1)/2=1992.
now n^2=1992*1992 so ans is 1992 means option 6.

Like? Yes | No   

THUMMALA VINEESHA   8 years AGO

1992
using formulaa of ap
1992[(1+3983)/2]*(1/1992)=1992

Like? Yes | No   

saitejapinky   8 years AGO

sum of first n odd numbers is
1+3+5+7+..........(2n-1) is n*n.
They given that 2n-1=3983
2n=3984
n=1992
1+3+5+.......3983=n*n
=1992*1992
but they given that (1+3+5+.......3983)/1992=(1992*1992)/1992
=1992.

Like? Yes | No   

Saravanakumar   8 years AGO

total numbers in the serious is 1992. addition of odd numbers = n^2 so 1992*1992/1992 =1992

Like? Yes | No   

Aamir Akhtar   8 years AGO

it is a arithmatic progression .. let n is the no of terms in this A.P.
nth term of A. P. = a + (n - 1)d = 3983
n = 1992
now sum of n terms of A. P. = n/ 2( 2a + ( n - 1 )d)
sum = 1992* 1992
so sum / 1992 = 1992
Ans = 1992

Like? Yes | No   

Jagapathi   8 years AGO

sum of n terms in a.p= n/2(a+l)
n=1992 a=1 l=3983
1992/2(3984)=1992*1992
1992*1992/1992=1992

Like? Yes | No   

neelesh   8 years AGO

for finding n use a+(n-1)d and for sum use n(first+last term)/2

Like? Yes | No   

PODETI SRINIVAS GOUD   8 years AGO

consider numerator 1+3+5+7+.......+3983
these are in arithmetic progression
where a(first term=a1)=1,d=a2-a1=3-1=2
nth term(sn)=a+(n-1)d=1+(n-1)2=2n-1
but given nth trem is 3983
2n-1=3983
2n=3984
n=1992
sum of 2 odd terms=1+3=4=2^2
sum of 3 odd tems=1+3+5=9=3^2
sum of 4 odd terms=1+3+5+7=16=4^2
.............................................................................
n tems sum of odd tems is n^2
here n=1992
n^2=1992^2
given eqn is
(1+3+5+7+............3983)/1992=1992^2/1992=1992

Like? Yes | No   

Jamil Ahmed   8 years AGO

Here no. of terms = (3983-1)/2 + 1 = 1992

Therefore, sum of all terms i.e. 1+3+5+7+.........+3983 = 1992/2*{2*1 + (1992-1)*2} = 1992*1992.

So, the answer is 1992*1992/1992 = 1992

Like? Yes | No   

suryakant singh   8 years AGO

total number of term is 1992,
sum of numbers is 1992/2(2+1991*2) = 1992/2(3984)
= 1992*1992
A/Q,(1992*1992)/1992 = 1992

Like? Yes | No   

sadhana   8 years AGO

sum of n odd numbers
1+3+5+.....+(2n-1)=n*n
so in the question
1+3+5+7+....(2(1992)-1)=1992*1992
so the ans....1992

Like? Yes | No   

UMANG RASTOGI   8 years AGO

by formula Sn = n/2(a+l) = n/2{2a+(n-1)d}
we get n=1992

so Sn/1992 =c 1992/2(1+3983) /1992 = 1992 ANS.

Like? Yes | No   

Jebas Gnana Prakash E   8 years AGO

Add 1 to the last no we get
3983+1=3984
Divide by 2 we get
3984/2=1992 //assume equation 1
Take
1992*1992=3968064
3968064/1992=1992 assume equation 2
Let
equation 1= equation2
hence ans is 1992

Like? Yes | No   

Mathi Vanan P   8 years AGO

add the last number with one =>(3983+1)=3984
divide the resultant value by 2=>(3984/2)=1992
square the resultant value=>(1992*1992)=3968064
in the question divide by 1992 is there therefore the resultant value is divided 1992 then
(3968064/1992)=1992

Like? Yes | No   

lingadurai   8 years AGO

3983+1=3984
3984/2=1992
1992*1992=3968064
3968064/1992=1992

Like? Yes | No   

Tanvi Jain   8 years AGO

Sum of first n odd no=n*n
=> 3983=1+(n-1)*2 (because d=2)
=> n=1992

For above question= 1992*1992/1992=1992

Like? Yes | No   

Surabhi Yadav   8 years AGO

1+3+5+7......+3983/1992
By Solving we get
1992*1992/1992
=1992

Like? Yes | No   

Vignesh S   8 years AGO

formula to find Sum
n/2(a+An)
To find n
an=a+(n-1)d
an=3983
a=1
d=2
3983=1+(n-1)2
3982/2=n-1
1991=n-1
n=1992
Sum (n/2)*(a+an)
(1992/2)*(1+3983)=====1992

Like? Yes | No   

anju sukumaran   8 years AGO

sum of AP= n/2(1st term + last term)......(1)
3983= 1+(n-1)2
=> n= 1992......(2)
substitute (2) into (1) to get sum of AP
=>sum of AP/1992 = 1992
hence ans=1992

Like? Yes | No   

Maheshwaran n   8 years AGO

(1+3983)/1992+(2+3982)/1992+......1992/1992
2+2+.....1=1992

Like? Yes | No   

Akshat Mokashi   8 years AGO

It is an AP with const. difference d=2
a=1 n=1991.5 and l=3983
so by applying formula for sum of n terms in AP we get
sum = (n/2)(a+l) = (1991.5/2)(1+3983) = (1991.5)(1992).
Now it is further divided by 1992 so ans is 1991.5 round to 1992

Like? Yes | No   

rajni   8 years AGO

Because Tn=a+(n-1)d ; from where we can have the value of n;i.e n=1992
and Sn=(first term + last term)n/2 ==> Sn/1992=1992

Like? Yes | No   

Akhilesh Singh Choudhary   8 years AGO

Numerator term is in A.P.
As we know that
a+(n-1)d=t
put the value of all to find the value of n first
i.e. 1+(n-1)2=3983
solved and We get n=1992
again we know that n/2(a+l)
where l=last term =3983
put the value of all and divide by 1992
we get the required ans i.e. 1992
option 6

Like? Yes | No   

Akshay   8 years AGO

it is an ap series so diff same so sum = n(a+l)/2
now put 3983=1+ (n-1)2
n= 1992
put n n get answer sum = 1992

Like? Yes | No   

GOURANGA SUNDAR MANDAL   8 years AGO

last term of the series is 3983...so l=3983
we know l=a+(n-1)*d..
so n=1992..
sum of the term sn=n/2{2a+(n-1)d}
sum =1992*1992...
so ans is 1992..
so option 6 is correct answer

Like? Yes | No   

Saravanakumar. R   8 years AGO

i got this answer by the shortened version of addition of odd numbers till 25.

Like? Yes | No   

CHHAYA   8 years AGO

it forms an airthmetic pprogression.
a(n)=3983; a=1 ;d=2
a(n)=a+(n-1)d;
3983=1+(n-1)2;
n=1992
S(n)=n/2(a+a(n));
S(n)=1992*1992;
S(n)/1992=1992;
so answer is 1992

Like? Yes | No   

aditya   8 years AGO

from 1 to 3983 there are 1992 numbers.
n=1992
(1992*1992)/1992=1992

Like? Yes | No   

ABID PASHA   8 years AGO

The sum of first n odd numbers=n^2
And the number of odd terms given by [1+n]/2
so we have
[1-3983]/1992=1992 terms

So

1992^2/1992=1992

Like? Yes | No   

Akanksha Varshney   8 years AGO

let the no of terms be n
3983=1+(n-1)×2
n=1992
sum of n odd consecutive terms=n^2
thus,ans=1992

Like? Yes | No   

nishant jain   8 years AGO

here it is an a.p
a=1 d= 2 l= 3983
therefore an= a + (n-1)d
n = 1992
Sn = n/2 * ( a+ l)
Sn = (1992*3984)/2
now ,Sn/1992
ans = 1992
option 6 is correct

Like? Yes | No   

Akshat Sood   8 years AGO

AP=n/2(2*a1+(n-1)*2
Average=sum of n terms/no of terms
n=n/3984(2*1+(n-1)*2
3984n=n(2+2n-2)
3984n=2n^2
n=1992

Like? Yes | No   

Saleel Kulkarni   8 years AGO

In the series the total numbers = n = (3983+1)/2 = 1992
a1=1
difference between two numbers =d = 2
Sum/1992=((n/2)(2*a1+(n-1)*d))/1992
=((1992/2)(2+3982))/1992
=1992
Option 6

Like? Yes | No   

bhavika   8 years AGO

Total odd nos from 0 to 3983 are 1992. Sum=1992*1992.hence answer is 1992*1992/1992=1992

Like? Yes | No   

J   8 years AGO

Sum of n terms in Arithmetic Progression is n/2(2a+(n-1)d)
So for the above task we get 1992*1992
when it is divided by 1992 remains as 1992.

Like? Yes | No   

Nikhil Shriwastav   8 years AGO

The sum of the first n odd numbers = n^2
And the number of odd terms is given by [1 + n] / 2
So we have[1 + 3983] / 2 = 1992 terms
So 1992^2 / 1992 =1992

Like? Yes | No   

Elamaruthu   8 years AGO

1+3+5+....+3983

take the last number 3983
and add 1 in that number 3984
3984/2=1992
and 1992*1992=3968064 is the sum answer
then (1992*1992)/1992=1992(ans)

Like? Yes | No   

swastik mukherjee   8 years AGO

(1+3+5+7+...)= (last terms +1)/2=no of terms
(3983+1)/2=1992
now n*n formula.
(1992*1992)/1992=1992

Like? Yes | No   

ranjitha   8 years AGO

By the arithmetic progress equation,
Tn=a+(n-1)d we will get number of terms, n= 1992
Then using equation,
Sn=(first term+last term)*n/2 and dividing the answer by 1992
We will get answer = 1992

Like? Yes | No   

TRAYAMBAK MISHRA   8 years AGO

s=(n/2)*(A1+An)
n is total number of term and A1 is first and An is last term.

Like? Yes | No   

S.MANJULA   8 years AGO

By AP tn =a+(n-1)d
So 3983 =1+(n-1)d
n =1992
Sum =(n/2)(1+3983 )
Sum =3968064
Sum/2=1992
Option 6

Like? Yes | No   

Saurav Jaiswal   8 years AGO

sum of series = 1992(3984)/2
so after dividing with 1992.we will get 1992

Like? Yes | No   

abhi   8 years AGO

last term=3983
3983=1+(n-1)*2
n=1992
sum=n/2*(a+l)
sum=1992/2*(1+3983)
total sum=(1992*(3984)/2)/1992
total sum=1992

Like? Yes | No   

Vaibhav Khedkar   8 years AGO

no of terms=1992 by the formula;
nth term= a+ (n-1)d
sum of n terms= n/2(2a+(n-1)d)
sum/1992=1992/2(2+1991*2)/1992
ans=1992
option(6)

Like? Yes | No   

Dinesh kumar sah   8 years AGO

n=3983-1/2+1=1992
and Sn=n/2(a+an)
Sn=1992/2(1+3983)
and in question
Sn/1992=1992 Ans

Like? Yes | No   

Akhil Kumar Arshid   8 years AGO

3983=1+(n-1)2
=> n=1992
sum = 1+3+5+.......+3983
sum=n*(a+l)/2
sum=1992*(1+3983)/2
total sum=(1+3+5+......+3983)/1992
total sum=((1992)*(3984)/2)/1992
=>total sum=1992
option 6

Like? Yes | No   

Nikhil Agarwal   8 years AGO

n = no. of terms in AP.
3983 = 1 + (n - 1)2
n = 1992.

now
(1 + 3+ 5+......+3983) = (n / 2)(1 + 3983)
(1 + 3+.........+3983) = (1992 *3984)/2
(1+3+5+......+3983)/1992 = 1992

Like? Yes | No   

Dhruv   8 years AGO

No. of terms in the series = 1992
Sum of n terms of AP = (n/2)(first term + last term)
= (1992/2)*(3984)
Sum/1992 = 3984/2 = 1992

Like? Yes | No (1)  

kavya   8 years AGO

numerator is 1992*1992
by using formulae (n(a+l))/2
n=1992,a=1,l=3983.

Like? Yes | No (1)