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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
Let $a, b, c$ be positive integers such that $\frac{b}{a}$ is an integer. If $a, b, c$ are in geometric progression and the arithmetic mean of $a, b, c$ is $b+2$, then the value of $\frac{a^2+a-14}{a+1}$ is
  • A
    $1$
  • B
    $2$
  • C
    $3$
  • ✓
    $4$
✓

Answer

Correct option: D.
$4$
d
Let $b=a r, c=a r^2 \Rightarrow r$ is Integers

Also $\frac{a+a r+a r^2}{3}=a r+2 \Rightarrow a+a r^2=2 a r+6$

$\Rightarrow \quad a ( r -1)^2=6 \quad \Rightarrow \quad r$ must be 2 and $a =6$.

Thus $\frac{a^2+a-14}{a+1}=\frac{36+6-14}{7}=4$ Ans.

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