Let a_1, a_2, a_3, .. be a sequence of positive integers in arith… - Vidyadip

MCQ

Let $a_1, a_2, a_3, \ldots .$. be a sequence of positive integers in arithmetic progression with common difference $2$. Also, let $b_1, b_2, b_3, \ldots .$. be a sequence of positive integers in geometric progression with common ratio $2$ . If $a_1=b_1=c$, then the number of all possible values of $c$, for which the equality

$2\left(a_1+a_2+\ldots .+a_n\right)=b_1+b_2+\ldots . .+b_n$

holds for some positive integer $n$, is. . . . . . .

✓
$1$
B
$5$
C
$8$
D
$7$

✓

Answer

Correct option: A.

$1$

a
Given $2\left( a _1+ a _2+\ldots . .+ a _{ n }\right)= b _1+ b _2+\ldots . .+ b _{ n }$

$\Rightarrow \quad 2 \times \frac{ n }{2}\left(2 c +( n -2) x _2\right)= c \left(\frac{2^{ n }-1}{2-1}\right)$

$\Rightarrow \quad 2 n ^2-2 n = c \left(2^{ n }-1-2 n \right)$

$\Rightarrow \quad c =\frac{2 n ^2-2 n }{2^{ n }-1-2 n } \in N$

$\text { So, } 2 n ^2-2 n \geq 2^{ n }-1-2 n$

$\Rightarrow \quad 2 n ^2+1 \geq 2^{ n } \Rightarrow n <7$

$\Rightarrow \quad n \text { can be } 1,2,3, \ldots.$

Checking $c$ against these values of $n$

we get $c=12 \quad($ when $n=3)$

Hence number of such $c=1$

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