12 + 13 + 14 + ... 51 + 52 + 53 = ?

Question

A. 1361
B. 1362
C. 1363
D. 1364
E. 1365

explanation showed adding in pairs which added to 65 and then y - x + 1 = 53 - 12 + 1 = 42
and then 21 pairs add to 65 and 65 x 21 and looked at the last digit which was 5

MY QUESTION : WHY IS IT 21 PAIRS? How would we know this?

KarishmaB · Accepted Answer

You can do this question in different ways:

Sum = 12 + 13 + 14 + ... 51 + 52 + 53

Method 1:

Sum of n consecutive positive integers starting from 1 is given as n(n+1)/2

Sum of first 53 positive integers = 53*54/2
Sum of first 11 positive integers = 11*12/2

Sum = 12 + 13 + 14 + ... 51 + 52 + 53 = 53*54/2 - 11*12/2 = 53*27 - 66
Note that 3*7 is 21 so 53*27 will end with 1 and when we subtract 66 from it, last digit will be 5. So answer must be (E)

Method 2:

From 12 to 53, how many numbers are there? 53 - 12 + 1 = 42 numbers
Number of positive integers from A to B are calculated as B - A + 1. You add an extra 1 to make up for the 12 that you subtracted. Since you want to include 12 as well in your counting, you add 1.

The average of these terms is (First term + last term)/2 = (53 + 12)/2 = 32.5
Average of a list is the number which can replace all numbers such that the sum of the list stays the same. So sum of the list can be calculated as Average*Number of elements in the list

Sum = 32.5*42 = 65*21 
Ends with 5 so answer (E)

RG800 · Answer

The sum is simply no. of terms multiplied by the average.

Number of terms:

53 - 11 = 42.

(This answers your question: No. of pairs = 42/2 = 21)

Average of each pair:

(53+12)/2 = 32.5
(52+13)/2 = 32.5
(51+14)/2 = 32.5
. .
. .

. .
So, total average = 32.5

Ans: 42*32.5 = 21*65

sagnik2422 · Answer

Where did 65 and 21 come from in this : Sum = 32.5*42 = 65*21

KarishmaB · Answer

Sum = 32.5 * 42 = 32.5 * 2 * 21 = 65 * 21 
We do this to remove the decimal and get integers.

KrishnakumarKA1 · Answer

Sum = sum(1 +2 +3…..53) - sum(1 +2+3….11) 
= 53(53 +1)/2 - 11(11 +1)/2 
= 1431- 66 
= 1365. Option E

AbhimanyuDhar · Answer

Well there are faster methods that have been posted by various members , this method shows one more possibility of how some seq can be solved.

how many terms are there between 12 and 53 ( inclusive) = 42
If we can see that 12 +13= 25 and then 14+15= 29 and 16+17 =33 So each next pair is 4 more than the previous pair 
How many pairs can we form then 21.

Sum = N ( FT + LT)/2
 21( 25 +105)/2= 21*65
( 20+1)* 65= 1300+65= 1365

georgioskonst · Answer

Simple :

(53+12/2) x (53-12+1)

MathRevolution · Answer

12 + 13 + 14 + ... 51 + 52 + 53 = (1 + 2 + 3 +...+ 53) - ( 1 + 2 + 3.... + 11)

Sum of 'n' natural numbers :  \frac{n (n + 1 ) }{ 2}

=> (1 + 2 + 3 +...+ 53) =  \frac{53 * 54 }{ 2}  = 1,431

=> (1 + 2 + 3 +...+ 11) =  \frac{11 * 12 }{ 2}  = 66

=> 1,431 - 66 = 1,365

Answer E

SIDDHARTH1901 · Answer

No. Of terms = 53-12+1= 42.

Sum = (1st term + last term)*(no. 
 of terms) /2
 = (12+53)*42/2
 = 1365.

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12 + 13 + 14 + ... 51 + 52 + 53 = ?

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