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  • Let <img alt="\mathrm{A=\left\{1, a_{1}, a_{2} \ldots \ldots a_{18} ,77\right\}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BA%3D%5Cleft%5C%7B1%2C%20a_%7B1%7D%2C%20a_%7B2%7D%2

Let \mathrm{A=\left\{1, a_{1}, a_{2} \ldots \ldots a_{18} ,77\right\}} be a set of integers with \mathrm{1<a_{1}<a_{2}<\ldots . \ldots a_{18}<77}. Let the set \mathrm{A+A=\{x+y: x, y \in A\}} contain exactly 39 elements. Then, the value of \mathrm{a_{1}+a_{2}+\ldots+a_{18}}  is equal to________.

Option: 1

702


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The minimum no. of different sums of 2 elements (not neccessarly distinct) at a set containg 20 distinct elements is 39, and it occurs ; if f the elements are in A. P

For e.g  \mathrm{Let\; B= \left \{ 1,2,3,\cdots 20 \right \}}

\mathrm{Min \; sum = 1+1= 2,\; Max \; sum= 20+20= 40}
So No.of diff sums \mathrm{= \left \{ 2,3,4,\cdots 40 \right \}= 39\; sums}

So \mathrm{a_{1},a_{2},\cdots a_{12}= 5,9,13,\cdots 73}
 So  \mathrm{a_{1},a_{2},\cdots a_{12}= \frac{18}{2}\left [ 5+73 \right ]= 702}

Posted by

Gautam harsolia

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