JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 14)
Let a1, a2, a3, .... be a sequence of positive integers in arithmetic progression with common difference 2. Also, let b1, b2, b3, .... be a sequence of positive integers in geometric progression with common ratio 2. If a1 = b1 = c, then the number of all possible values of c, for which the equality 2(a1 + a2 + ... + an) = b1 + b2 + ... + bn holds for some positive integer n, is ...........
Answer
1
Explanation
Given arithmetic progression of positive integers terms a1, a2, a3, ..... having common difference '2' and geometric progression of positive integers
terms b1, b2, b3, .... having common ratio '2' with a1 = b1 = c, such that
2(a1 + a2 + a3 + ... + an) = b1 + b2 + b3 + ... + bn
$$ \Rightarrow 2 \times {n \over 2}[2C + (n - 1)2] = C\left( {{{{2^n} - 1} \over {2 - 1}}} \right)$$
$$ \Rightarrow 2nC + 2{n^2} - 2n = {2^n}.C - C$$
$$ \Rightarrow C[{2^n} - 2n - 1] = 2{n^2} - 2n$$
$$ \because $$ $$C \in N \Rightarrow 2{n^2} - 2n \ge {2^n} - 2n - 1$$
$$ \Rightarrow 2{n^2} + 1 \ge {2^n} \Rightarrow n \le 6$$
and, also C > 0 $$ \Rightarrow $$ n > 2
$$ \therefore $$ The possible values of n are 3, 4, 5, 6
So, at $$n = 3,\,C = {{(2 \times 9) - 6} \over {8 - 6 - 1}} = 12$$
at, $$n = 4,\,C = {{32 - 8} \over {16 - 8 - 1}} = {{24} \over 9} = {8 \over 3} \notin N$$
at, $$n = 5,\,C = {{50 - 10} \over {32 - 10 - 1}} = {{40} \over {21}} \notin N$$
and at, $$n = 6,\,C = {{72 - 12} \over {64 - 12 - 1}} = {{60} \over {51}} \notin N$$
$$ \therefore $$ The required value of C = 12 for n = 3
so number of possible value of C is 1
terms b1, b2, b3, .... having common ratio '2' with a1 = b1 = c, such that
2(a1 + a2 + a3 + ... + an) = b1 + b2 + b3 + ... + bn
$$ \Rightarrow 2 \times {n \over 2}[2C + (n - 1)2] = C\left( {{{{2^n} - 1} \over {2 - 1}}} \right)$$
$$ \Rightarrow 2nC + 2{n^2} - 2n = {2^n}.C - C$$
$$ \Rightarrow C[{2^n} - 2n - 1] = 2{n^2} - 2n$$
$$ \because $$ $$C \in N \Rightarrow 2{n^2} - 2n \ge {2^n} - 2n - 1$$
$$ \Rightarrow 2{n^2} + 1 \ge {2^n} \Rightarrow n \le 6$$
and, also C > 0 $$ \Rightarrow $$ n > 2
$$ \therefore $$ The possible values of n are 3, 4, 5, 6
So, at $$n = 3,\,C = {{(2 \times 9) - 6} \over {8 - 6 - 1}} = 12$$
at, $$n = 4,\,C = {{32 - 8} \over {16 - 8 - 1}} = {{24} \over 9} = {8 \over 3} \notin N$$
at, $$n = 5,\,C = {{50 - 10} \over {32 - 10 - 1}} = {{40} \over {21}} \notin N$$
and at, $$n = 6,\,C = {{72 - 12} \over {64 - 12 - 1}} = {{60} \over {51}} \notin N$$
$$ \therefore $$ The required value of C = 12 for n = 3
so number of possible value of C is 1
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