$a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are first five terms of an A.P. such that $a_{1} + a_{3} + a_{5} = - 12$ and $a_{1} a_{2} a_{3} = 8$ . Find the first term and the common difference.

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Solution

Let $d$ be the common difference of AP
Given $a_{1} + a_{3} + a_{5} = - 12 a_{1} + a_{1} + 2 d + a_{1} + 4 d = - 12 3 a_{1} + 6 d = - 12$
$a_{3} = a_{1} + 2 d = - 4 \Rightarrow d = \frac{- 4 - a_{1}}{2} (1)$
Given, $a_{3} = - 4$
$a_{1} a_{2} a_{3} = 8 \Rightarrow a_{1} a_{2} (- 4) = 8 \Rightarrow a_{1} a_{2} = - 2 a_{1} + (a_{1} + d) = - 2$
From $(1)$
$a_{1} (a_{1} + \frac{(- 4 - a_{1})}{2}) = - 2$
$a_{1} (a_{1} - 4) = - 4 \to$ ${a_{1}}^{2} - 4 a_{1} + 4 = 0 (a_{1} - 2)^{2} = 0 a_{1} = 2 (2)$
From $(1), (2)$
$d = \frac{- 4 - a_{1}}{2} = \frac{- 4 - 2}{2} = - 3 a_{1} = 2, d = - 3$

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Similar questions

a_{1}, a_{2}, a_{3}, . . . . . . . a_{n}

a_{1} = 0

d \neq 0

\frac{a_{3} - a_{2}}{a_{2}} + \frac{a_{4} - a_{2}}{a_{3}} + \frac{a_{5} - a_{2}}{a_{4}} . . . . . . . . . . . . \frac{a_{n} - a_{2}}{a_{n - 1}}

a_{1}, a_{2}, a_{3} . . . . . . . . . a_{n},

a_{1} + a_{3} + a_{5} = - 12

a_{1} a_{2} a_{3} = 8

a_{2} + a_{4} + a_{6}

A . P .

a_{1}, a_{2}, a_{3}, . . . . ., a_{n}, . . . ., a_{1} + a_{3} + a_{5} = - 12

a_{1} a_{2} a_{3} = 8

a_{2} + a_{4} + a_{6}

A . P .

a_{1}, a_{2}, a_{3}, . . . . . ., a_{n}, . . . . .

a_{1} + a_{3} + a_{5} = - 12

a_{1} a_{2} a_{3} = 8

a_{2} + a_{4} + a_{6}

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