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Evaluation of a Determinant

Let a1,a2,......

Question

Let $a_{1}, a_{2}, . . ., a_{10}$ be an AP with common difference $- 3$ and $b_{1}, b_{2}, . . ., b_{10}$ be a GP with common ratio $2.$ Let $c_{k} = a_{k} + b_{k}, k = 1, 2, . . ., 10.$ If $c_{2} = 12$ and $c_{3} = 13,$ then $10 \sum k = 1 c_{k}$ is equal to

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Solution

$c_{2} = a_{2} + b_{2} = (a_{1} - 3) + 2 b_{1} = 12 \Rightarrow a_{1} = 11$
$c_{3} = a_{3} + b_{3} = (a_{1} - 6) + 4 b_{1} = 13 \Rightarrow b_{1} = 2$
$c_{k} = a_{k} + b_{k} = (a_{1} - 3 (k - 1)) + (b_{1} \cdot 2^{k - 1})$
$= (11 - 3 k + 3) + (2^{k}) = 14 - 3 k + 2^{k}$
$10 \sum k = 1 c_{k} = 10 \sum k = 1 (2^{k} - 3 k + 14)$
$= 10 \sum k = 1 2^{k} - 3 10 \sum k = 1 k + 10 \sum k = 1 14$
$= 2 (2^{10} - 1) - 3 \cdot \frac{10 \cdot 11}{2} + 140 = 2021$

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Similar questions

b_{i} > 1

for i = 1, 2, .... ,101. Suppose

l o g_{e} b_{1}, l o g_{e} b_{2}, \dots \dots l o g_{e} b_{101}

are in AP with the common difference

l o g_{e} 2

a_{1}, a_{2}, \dots \dots a_{101}

are in AP, such that

a_{1} = b_{1}

a_{51} = b_{51}

t = b_{1} + b_{2} + \dots \dots + b_{51}

s = a_{1} + a_{2} + \dots \dots + a_{51}

a_{1}, a_{2}, a_{3}, \dots

\frac{a_{1} + a_{2} + \dots + a_{10}}{a_{1} + a_{2} + \dots + a_{p}} = \frac{100}{p^{2}},

p \neq 10,

\frac{a_{11}}{a_{10}}

C_{k} =

^{n} C_{k}

0 \leq k \leq n

A_{k} = [\begin{matrix} C_{k - 1}^{2} & 0 0 & C_{k}^{2} \end{matrix}]

k \geq 1

A_{1} + A_{2} + . . . + A_{n} = [\begin{matrix} k_{1} & 0 0 & k_{2} \end{matrix}]

a_{1}, b_{1}, c_{1}

be natural numbers. We define

a_{2} = g c d (b_{1}, c_{1}), b_{2} = g c d (c_{1}, a_{1}), c_{2} = g c d (a_{1}, b_{1}),

a_{3} = l c m (b_{2}, c_{2}), b_{3} = l c m (c_{2}, a_{2}), c_{3} = l c m (a_{2}, b_{2}) .

g c d (b_{3}, c_{3}) = a_{2}

Q. If each pair of equations

a_{1} x^{2} + b_{1} x + c_{1} = 0, a_{2} x^{2} + b_{2} x + c_{2} = 0

a_{3} x^{2} + b_{3} x + c_{3} = 0

has a common root, then show that
(i)

\frac{c_{1} a_{2} + c_{2} a_{1}}{c_{1} a_{2} - c_{2} a_{1}} + \frac{a_{1} a_{2} b_{3}}{a_{3} (a_{1} b_{2} - a_{2} b_{1})} = 0

{(\frac{a_{1} b_{2} - a_{2} b_{1}}{a_{1} c_{2} - a_{2} c_{1}})}^{2} = \frac{a_{1} a_{2} c_{3}}{c_{1} c_{2} a_{3}}

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