10 + 11 + 12 +...+ x = 5106 What is x?

Method-1:
Here are some properties that are useful for people to know.



Now, Sum of all the terms in any Arithmetic Progression \(= (FirstTerm+LastTerm)*(\frac{No.ofTerms}{2})\)

So, 10 + 11 + 12 +...+ x = 5106 (which has total terms = x-10+1 = x-9 terms) becomes
\(10 + 11 + 12 +...+ x = (10+x)*\frac{(x-9)}{2} = 5106\)

i.e. \((x-9)*(x+10) = 2*5106 = 2*46*111 = 92*111\)

i.e. x-9 = 92
i.e. x = 101

Method-2: USE Options

Number of terms in series 10 to x will be = x-10+1 = x-9

i.e. x-9 must be a factor of 2*5106

\(5106 = 2*3*23*37\)

Using options

A. 100 i.e. x-9 = 100-9 = 91 which is NOT factor of 2*5106
B. 101 i.e. x-9 = 101-9 = 92 = 2*2*23 which is a factor of 2*5106
C. 102 i.e. x-9 = 102-9 = 93 which is NOT factor of 2*5106
D. 103 i.e. x-9 = 103-9 = 94 which is NOT factor of 2*5106
E. 104 i.e. x-9 = 104-9 = 95 which is NOT factor of 2*5106

Answer: Option B

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Bunuel: Please shift this question to PS thread

Sum of first 100 terms =101*100 /2 = 5050

First 9 terms are missing = 9* 10/2 =45

Total =5050-45 =5005

in order to make it 5160 we need to add 101

Therefore IMO B

Sum of first x natural number = x(x+1)/2 = 5106+45
Where 45 is the sum of the first 9 numbers (which is missing in the series)

X(X+1) = 10302

must be 101,102 or 103,104 (last digit is 2)
quickly solving the pairs helps to identify the answer
x = 101

We have a direct formula for sum of n terms in Arithmetic progression.
Sum of n terms= (n/2)*(2a+(n-1)d)

Where n is the number of terms in A.P
"a" is the first term in the series
"d" common difference between two consecutive terms.

No add 1-9 numbers on both sides so we get a new arithmetic series as 1+2+3+4+...+x =5151

Upon substituting the above values and making it equal to 5151 leaves the answer as x=101
(a=1, d=1, Sum =5151)

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