JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Sequences and Series

Let

A = {1, a_{1}, a_{2}, \dots, a_{18}, 77}

be a set of integers with

1 < a_{1} < a_{2} < \dots < a_{18} < 77

. Let the set

A + A = {x + y : x, y \in A}

contain exactly 39 elements. Then, the value of

a_{1} + a_{2} + \dots + a_{18}

is equal to ____.

Answer: 702

Solution

The set A has 20 elements. If all elements were distinct,

A + A

would have at most

(2 20) + 20 = 210

elements (all pairwise sums plus doubles).

For

A + A

to have exactly 39 elements, the elements of A must form an arithmetic progression. The minimum sum is

1 + 1 = 2

and maximum is

77 + 77 = 154

. With 39 distinct sums, we need the common difference to create this structure.

If A is an AP with first term 1, last term 77, and 20 terms total, the common difference is:

d = \frac{77 - 1}{19} = 4

A = {1, 5, 9, 13, \dots, 73, 77}

.

The sum

a_{1} + a_{2} + \dots + a_{18} = 5 + 9 + 13 + \dots + 73

(18 terms starting from 5):

= \frac{18 ( 5 + 73 )}{2} = \frac{18 \times 78}{2} = 702

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Sequences and Series

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