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Fallacy

Let a1, a2,...

Question

Let $a_{1}, a_{2}, a_{3}, . . . . . . . . . . . . . .$ be in A.P. and $q_{1}, q_{2}, q_{3}, . . . . . . . . .$ be in G.P. such that $a_{1} = q_{1} = 2$ and $a_{10} = q_{10} = 3$ then:

A

a_{1} q_{19}

is not an integer

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B

a_{19} q_{7}

is an integer

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C

a_{7} a_{18} = a_{19} q_{10}

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D

None

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Solution

The correct option is B $a_{7} a_{18} = a_{19} q_{10}$
$a_{10} = a_{1} + 9 d = 2 + 9 d = 3 ∴ d = 1 / 9$ ...................(1)
$q_{10} = q_{1} r^{9} = 2 r^{9} = 3 ∴ r^{9} = 3 / 2$ ............(2)
Now $a_{7} q_{19} = (2 + 6 d) 2 r^{18} = (2 + \frac{6}{9}) .2 \frac{9}{4}$
= $9 + 3 = 12$ by (1) and (2) .............(3)
Above is an integer. Hence (a) is ruled out
$a_{19} q_{10} = (2 + 18 d) 2 r^{6} = (2 + 18. \frac{1}{9}) .2 . \frac{3}{2}$
= $4.3 = 12$ by (1) and (2)
= $a_{7} q_{19}$ by (3)
Therefore, $a_{7} a_{18} = a_{19} q_{10}$

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