IIT-JEE 2002 ScreeningQuestion 1
Suppose $a,b,c$ are in A.P. and $a^{2},b^{2},c^{2}$ are in G.P. If $a<b<c$ and $a+b+c=\dfrac32$, then the value of $a$ is
Sequence and Series · Mixed Progressions (AP-GP)
Solution
Formulas used. (1) $a,b,c$ in A.P. $\Rightarrow a+c=2b$(2) $a^2,b^2,c^2$ in G.P. $\Rightarrow ac=\pm b^2$Step 1 - Middle term. $3b=\dfrac32$, so $b=\dfrac12$ and $a+c=1$. Step 2 - Product. $ac=\pm\dfrac14$. $ac=\dfrac14$ with $a+c=1$ gives $a=c=\tfrac12$ (rejected). So $ac=-\dfrac14$. Step 3 - Solve. $t^2-t-\dfrac14=0$: $$t=\frac{1\pm\sqrt2}{2}.$$ $a$ is the smaller root: $a=\dfrac12-\dfrac1{\sqrt2}$. Answer: (D)