JEE Advanced 2014 · Paper 1 · Mathematics Q51

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MathematicsInteger+3Question 51 of 60

Let $a,b,c$ be positive integers such that $\dfrac ba$ 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

Your answer

Answer and worked solution

Official answer: 4

GP. Write $b=ar$ , $c=ar^2$ with $r=b/a$ a positive integer.

Mean condition.
$\frac{a+ar+ar^2}3=ar+2\implies a(r^2-2r+1)=6\implies a(r-1)^2=6.$
The only square dividing 6 is 1, so $(r-1)^2=1$ ; $r$ is positive, so $r=2$ and $a=6$ (then $b=12$ , $c=24$ ; the mean 14 is $b+2$ ).

Value.
$\frac{36+6-14}{7}=\frac{28}7=4.$

Answer: 4

Topics:Geometric progression Arithmetic progression

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JEE Advanced 2014 · Paper 1, Mathematics question 51