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Arithmetic Progressions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsArithmetic Progressions2 Marks
Question
A sequence is defined by $\text{a}_\text{n}=\text{n}^3-6\text{n}^2-11\text{n}-6,\text{n}\in\text{N.}$ show that the first three terms of the sequence are zero and all other terms are positive.
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Answer

$\text{a}_\text{n}=\text{n}^3-6\text{n}^2-11\text{n}-6,\text{n}\in\text{N.}$The first three terms are $​​\text{a}_1,\text{ a}_2$ and $​​\text{a}_3$
$​​\text{a}_1=(1)^3-(1)^2+11(1)-6=0$
$​​\text{a}_2=(2)^3=(2)^2+11(2)-6=0$
$​​\text{a}_3(3)^2-(3)^2+11(3)-6=0$
$\therefore$ the $1^​​\text{st}\ 3$ terms are zero.
and
$\text{a}_\text{n}=\text{n}^3-6\text{n}^2-11\text{n}-6$
$=(​​​\text{n}​-2)^3-(\text{n}-2)$ is positive.

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