If (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280, then what

Question

If (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280, then what is the value of 1 + 2 + 3 + ... + n ?

A. 4850
B. 4851
C. 4852
D. 4853
E. 4854

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GMATWhizTeam · Accepted Answer

We need to find the value of  1 + 2 + 3 + ….n = \frac{n*(n + 1)}{2}

We know that:
 (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280 
 • Note that the numbers 1, 3, 5, ….. 47 are in arithmetic progression and the common difference is 2.
 o Using the above observation, we can find the total number of terms.  
   ⟹ 47 = 1 + (k – 1) * 2 
 ⟹ k = \frac{46}{2} + 1 = 24

Thus, the given expression can be written as:

(2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280 
 ⟹ 2n * 24 + (1 + 3 + 5…. + 47) = 5280 
 ⟹ 48 * n + \frac{24}{2} *   = 5280 
 ⟹ 48n + 12 * 48 = 5280

Dividing all sides by 48, we get:

⟹ n + 12 = 110 
 ⟹ n = 98

Thus, the sum of  1 + 2 + 3 + ….n = \frac{n*(n + 1)}{2} = 98 * \frac{99}{2} = 49 * 99 = 49 * (100 -1) = 4851

The correct answer is  Option B

Chethan92 · Answer

(2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280

Rearranging the terms, 1+3+...+47.
a = 1
d = 2
l = 47
a+(n-1)d = 47
n = 24

2n*24+Sum of 24 terms = 48n+n/2  = 48n+12(48) = 48n+576 = 5280
n = 98
Now, 1+2+3+....98 = n(n+1)/2 = 98*99/2 = 49*99 = (50-1)(100-1) = 5000-50-100+1 = 4851

B is the answer.

nick1816 · Answer

(n+24)^2 - n^2 = 5280 
 n^2+576 +48n -n^2 = 5280

48n= 48*(110-12)
n= 98

1 + 2 + 3 + ... + 98 = 98*99/2 (no need to calculate...unit digit must be 1)

B

hiranmay · Answer

If (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280, then what is the value of 1 + 2 + 3 + ... + n ?

A. 4850
B. 4851 -->  correct 
C. 4852
D. 4853
E. 4854

Solution:

r th term of the sequence = Tr = 2n+47 = a+(r-1)*d=2n+1+(r-1)*2 =2n+2r-1

=> Tr= 2n+2r-1 =2n+47
=> r=24, so total 24 terms in the sequence

sum of all the 24 terms = 24*2n+(2*24*25/2)-24=5280
=> 2n+25-1=220
=> 2n=196
=> n=98

so 1 + 2 + 3 + ... + n =98*99/2=49*99=49(100-1)=4900-49=4851

preetamsaha · Answer

(2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280

or,2n*24+(1+3+5+...+47) = 5280
or,48n+n/2  = 5280  
or,48n+12(48) = 5280
or,48n+576 = 5280 
or,n = 98
Now, 1+2+3+....98 = n(n+1)/2 = 98*99/2 = 49*99 = (50-1)(100-1) = 5000-50-100+1 = 4851

correct answer B

preetamsaha · Answer

nick1816

(n+24)2−n2=5280 => how did u establish this ?

nick1816 · Answer

1+3+5+7+.....+(2n+47) 
Sum of first n+24 odd numbers = (n+24)^2......(1)

1+3+5+7+.....+(2n-1) 
Sum of first n odd numbers = n^2....(2)

Subtract 2 from 1

we get 
(2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = (n+24)^2 - n^2

chetan2u · Answer

The moment I saw this question and the choices, I realized I do not require to find n to get the answer.

Sum of first n natural numbers =  \frac{n(n+1)}{2}=choice......n(n+1)=2*choice 
 So when I multiply the choices with 2, our solution should give us product of two integers..

A.  4850....4850*2=9700=97*100. ..The numbers in the product cannot be closer than this
Actually this should tell you that your answer is 98*99
B.  4851......4851*2=9702=98*99. ..

Now all choices are consecutive so We can be 100% sure that only ONE will be product of two consecutive integers.

B

Not that such ways will help you in all questions, but will surely save some precious time in 2-3 odd questions.

fireagablast · Answer

47-1/2 + 1 = 24 ---> number of terms original equation

5280 = 24/2 
5280 = 12 
440 = 4n+48
392 = 4n
98=n

1+2+3+4...98 = 98(98+1)/2 = 49*99 = 4851

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If (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280, then what

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If (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280, then what

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