1 + 3 + 5 + 7 + .......... + x = 12544 s is last term of the sequence. x will be

Surit Datta   9 years AGO

According to the summation series formula,
Sn = n/2*(2*a1 + (n - 1)*d)
where a1 is the first term of the sequence,
d is the common difference,
n is the number of the term to find.
Sn is the total summation,
we get after solving that,
n = 112
so according to the formula,
an = a1 + (n - 1)*d
putting the value of n =112
we get ,
an = 223
so option 7
that's it.

Like? Yes (9) | No (2)  

Sanjoy   9 years AGO

Sum of the series Sn =12544 ,
d= 2, and a is the 1st term
let x is the nth term of the series.
So we know, Sn =n*{2a+(n-1)*d}/2
12544=n*{2*1+(n-1)*2}/2
Or,12544=n*n
Or,n2=12544
Hence,n=112
x= nth term= a+(n-1)d
= 112th term=1+(112-1)*2
=223

Like? Yes (5) | No (2)  

Supriya Mondal   9 years AGO

We know that for A.P series
Sn=n/2[2*a+(n-1)d]
Sn=n/2[a+l]
where, Sn=Sum of n terms
n=number of terms
a=first term of the series
l=last term of the series
d=difference between two terms
Here,
Sn=12544
n=?
a=1
l=?
d=2
Putting this values on the first equation we get n=112
then putting this value on the second equation we get l=223
So, option 7)223 is the right answer

Like? Yes (6) | No (3)  

Anuj Rai   9 years AGO

Using s=(n*(a+l))/2

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Yathiraj   9 years AGO

Sum of N numbers => N^2
Nth number in series => N*2-1 => x in this case
Here is the sum is 12544 = N^2
so, N = 112
x = N*2-1 = 112*2-1 = 223

Option 7.

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s.durgadevi   9 years AGO

sn=n/a[2a+(n-1)d]
sn=12544
n=112
x=n
x=a+(n-1)d
x=223
option 7

Like? Yes (4) | No (1)  

PUJA MUKHERJEE   9 years AGO

n/2*(2+(n-1)*2)=12544
n=112
x=1+(112-1)*2
=223

Like? Yes (3) | No   

saurabh goel   9 years AGO

first calculate sn = n/2(a+l)
where n= 112
then cal x= a +(n-1)d where a=1,d=2,n=112

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sanu kumar verma   9 years AGO

at first we find the sum of infinite series by using formula sn=n/2(2a+(n-1)d.then we get n=112 put this value in this series when first tem and sum of infinite series are known i.e sn=n/2(a+l) and from here we get l=223

Like? Yes (2) | No   

Madhumoy Gope   9 years AGO

If x is the nth term of the series then,
x=1+(n-1)*2
Now,
n*(1+x)/2=12544
or,n*(1+1+(n-1)*2)/2=12544
or,n=112
So,the value of x=1+(112-1)*2=223

Like? Yes (3) | No (1)  

abhinav upadhyay   9 years AGO

n/2 [ 2+(n-1)2 ] = 12544
=> n/2 ( 2n ) = 12544
=> n=112
so , x = 1+ ( 112-1 ) 2
=> x = 223 ans-

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SUSHANT KUMAR   9 years AGO

1+3+5+7+....x=12544
so diff=2
sum=12544
total no of terms=112
hence last term=223

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sompon temtheerakij   9 years AGO

let y = numbers of the sequence
so:- (1+x)y/2 = 12544.....eq.1
(1+x)/2 = y .....eq.2
by solving 2 equation --> x = 223

Like? Yes (1) | No   

ABHINAV PUSHPAM   9 years AGO

Question is wrong. Instead of s, there should be x.
All should get 20 points
Although option 7) is correct.

Like? Yes (2) | No (1)  

vibhu raj gupta   9 years AGO

sum of AP term's=n/2(2a+(n-a)d)
=12544 by solving it
n=112
so series starts from 1
i.e. x=223

Like? Yes (1) | No   

shabana banu shaik   9 years AGO

by substituting in nth term formula and sum of n terms formula i.e a+(n-1)d, n value is (x+1)/2, putting n value in n/2(a+l)..... we get (x+1)^2/4 get (x+1)^2=12544*4=50176, we get 1+x=224, x=223

Like? Yes (2) | No (1)  

vivek   9 years AGO

As the sum of consecutive odd number results in square of total number of terms.
i.e 1+3+5=9 = (3^2)
1+3+5+7+9=25 = (5^2)....so
we have,
1+3+5+77+9+.....+x=12544
so, using above process number of terms= sqrt(12544)=112
Total number of terms in this series is 112.
now,
x=t(112)=a+(n-1)d (first term=1 and common diff.=2)
x=1+(112-1)2
x=1+(111*2)
x=1+222
so, x=223. ans

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NESAB JAFAR   9 years AGO

223+1=224
224/2=112
112*112=12544

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Anuj Rai   9 years AGO

Using ap formula
s=n/2(a+(n-1)d)

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Srijan   9 years AGO

n/2[2*a+(n-1)]d=12544; a=1,d=2 {sum of n terms in Arithmetic Progression}
=>we get n=112
last term=> Tn=a+(n-1)d
a=1,n=112,d=2
the last term 'x'=223

Like? Yes (1) | No   

shreya saklecha   9 years AGO

Sn=n/2[2a+(n-1)d];
12544=n/2[2(1)+(n-1)2];
n^2=12544;
n=112;

Sn=n/2[a+l];
12544=112/2[2+l];
l=223

Like? Yes (2) | No (1)  

siddharth prakash singh   9 years AGO

1 + 3 + 5 + 7 + .......... + x = 12544
it is an AP with 1st term 1 and common difference 2 and sum=12544.
using sum formula we get n=112.
so last term=223

Like? Yes (2) | No (1)  

Akhila H.K   9 years AGO

The answer is option 7)
Because the sum of the expression is 12544 ,
First we need to find out number of terms in AP,as this expression is in AP.
So,Sn=n/2[2a+(n-1)d]
12544=n/2[2*1+(n-1)2]
25088=n/2[2+2n-2] (here cancel 2)
Therefore 2n^2=25088
n^2=12544
so n=112

Then substitute n in
Tn=a+(n-1)d
=1+(112-1)2
Tn =223.
Hence the last term is x
So x=223.

Like? Yes (1) | No   

Devendra Marghade   9 years AGO

Sum of total 'n' terms of series 1,3,5,7,.....,x
=(n/2)[2*1+(n-1)*2]=12544, n^2=12544 or n=112
So, the last term 'x'=1+(112-1)*2=223

Like? Yes (2) | No (1)  

C.DEEPAK KUMAR   9 years AGO

12544=n/2(first +last terms)
12544=n/2(2*1+(n-1)2)
n2=12544
n=112
And x=223

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MRITUNJAY   9 years AGO

12544=n/2[2*1+(n-1)2]
25088=n*2n
n^2=12544
n=112
12544=112/2(1+x)
x=223

Like? Yes (2) | No (1)  

umesh   9 years AGO

sn=n/2(2a+(n-1)d)
so we get n=112
then n=(l-a)/d+1
so we get x=223

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Ashutosh kumar mishra   9 years AGO

sum =n/2(2*first_term+(n-1)*d)
calculate the values of n where
sum=12544
d=2 (common ratio)
first_term=1

so n=112
then
x=first_term+(n-1)*d
x=223

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DEBMALYA GHOSAL   9 years AGO

s/2(2*a1 +(s-1)*2)=12544
=> s=112

x= a1+(s-1)*d
=>1+(112-1)*2
=>223

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Pargat Pal Singh   9 years AGO

number of terms=112
x= 1+2*(112-1)=223

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ankit katiyar   9 years AGO

x=1+(n-1)*2
n=(x+1)/2


12544=n/2*(1+x)

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PHARATADE SANDIP NAMADEV   9 years AGO

underroot of 12544 is 112.
series have 112*2-1=223 trms

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SHAHEEN   9 years AGO

given series is an arithemetic progression with common difference=2
sum to n terms in AP= (n/2)[2(1)+(n-1)2]=12544
===>>n^2=12544
===>n=112
Also sum=12544=(n/2)[first term+last term]
===>12544=(112/2)[1+x]
......x=223
so option is 7

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alety kiranmai   9 years AGO

It is in AP with common difference=2
12544=(n/2)[2+(n-1)2]
n=112
from this 112th term=223
so option is 7

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aman aditya   9 years AGO

Sum of first 61 term is 12321
so x= 12544-12321 =223

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sangrambadi sahoo   9 years AGO

here in the ap a=1,d=2 so we can count sum by Sn=n/2(2a+(n-1)d)
here we get n so
tn= a+(n-1)d
we get tn= 223

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Niharika Singh   9 years AGO

n/2(2*a(n-1)d)=12544
solving this we get n=112
tn=a+(n-1)d
1+111*2=223

Like? Yes (1) | No (1)  

sravani   9 years AGO

sum is n(2a+(n-1)d)/2=12544
a is 1
d is 2
so, x is 223

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Amarnath Kumar Upadhyay   9 years AGO

sum of nth terms of A.p=n/2(2*a+(n-1)d)
so
12544=n/2(2*a+(n-1)d)
put a=1 and d=2
n=112
x=a+(n-1)d
x=223

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KUMAR HARSH   9 years AGO

let there are n term
x=1+(n-1)*2
x=2*n-1
12544=n/2[1+x]
12544=n/2[1+2*n-1]=n^2
n=112
x= 2*n-1=2*112-1=224-1=223

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Santhosh R   9 years AGO

Sn=n/2[2a+(n-1)d]=s /2[2(1)+(s-1)2]=12544
s=112
Tn =a+(n-1)d=1+(112-1)2
hence Tn =x=223

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PAWAN KUMAR   9 years AGO

Total no of elements = n
So, x= 2n-1
n(1+x)/2 = 12544
=> n = 112
=> x =112*2 - 1 = 223. Ans

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neha   9 years AGO

sum of ap= n/2(first term +last term)
nth term of an ap= a+(n-1)d [whre a=first trm, d= comon differcne, n=no. of terms]

so nth term=> x= 1+(n-1)*2=>
x=(2n-1) ---------first eqn
and sum=>12544=(n/2(1+x)) ---------- second eqn
put n from first eqn in second eqn

12544=(x+1)(x+1)/2
12544*4=(x+1)^2
x+1=224
x=223
ans.

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Meenakshi   9 years AGO

The given series is A.P
d=2,a=1
Tn =a+(n-1)d
=1+(n-1)2
Tn =2n-1
Sum of series=n(a+l)/2
12544=n(1+2n-1)/2 (here l is the last term)
12544=n2
112=n
So last term=2(112)-1
=223
Answer is option 7)

Like? Yes (1) | No (1)  

sourabh verma   9 years AGO

sum of the no's in a.p is
s=n/2[2a+(n-1)*d]
so
12544=n/2[2+(n-1)2]
we will get n=112
as x is the last term so
x=a+(n-1)*d
so x will be 223

Like? Yes (1) | No (1)