If a, b, c are in A.P. and a2, b2, c2 are in G.P. such that a < b < c and a + b + c = 3 \over 4, then the value of a is :

Sequences and Series PYQ — JEE Main 2018 · 15 Apr · Shift 2 · Q23 with Solution

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Correct answer: C

Since $a,b,c$ are in A.P., we can write

\qquad c=a+2d$$ for some $d>0$ (because $a<b<c$). Also, $$a+b+c=\frac34$$ so $$a+(a+d)+(a+2d)=\frac34$$ $$3a+3d=\frac34$$ $$a+d=\frac14$$ Hence, $$b=\frac14.$$ So the three numbers are $$a=\frac14-d,\qquad b=\frac14,\qquad c=\frac14+d.$$ --- 2. **Use the G.P. condition on squares** Given that $a^2,b^2,c^2$ are in G.P., we have $$b^4=a^2c^2.$$ Since $a<b<c$ and $a+b+c=\frac34>0$, here all values come out positive, so we may also use $$b^2=ac.$$ Now substitute: $$\left(\frac14\right)^2=\left(\frac14-d\right)\left(\frac14+d\right).$$ Thus, $$\frac1{16}=\frac1{16}-d^2$$ which gives $$d^2=0.$$ That would force $a=b=c$, contradicting $a<b<c$. So let us carefully use the actual G.P. condition: $$\frac{b^2}{a^2}=\frac{c^2}{b^2}$$ $$b^4=a^2c^2$$ $$b^2=|ac|.$$ Because $a<b<c$ and one of them can be negative, we should allow $$ac=-b^2.$$ Now substitute: $$\left(\frac14-d\right)\left(\frac14+d\right)=-\left(\frac14\right)^2$$ $$\frac1{16}-d^2=-\frac1{16}$$ $$d^2=\frac18$$ $$d=\frac1{2\sqrt2}$$ (since $d>0$). Therefore, $$a=\frac14-d=\frac14-\frac1{2\sqrt2}.$$ --- 3. **Match with the options** This is exactly $$\boxed{\frac14-\frac1{2\sqrt2}}$$ which is **Option C**. --- 4. **Verification with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

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Correct answer: C

Since $a,b,c$ are in A.P., we can write

\qquad c=a+2d$$ for some $d>0$ (because $a<b<c$). Also, $$a+b+c=\frac34$$ so $$a+(a+d)+(a+2d)=\frac34$$ $$3a+3d=\frac34$$ $$a+d=\frac14$$ Hence, $$b=\frac14.$$ So the three numbers are $$a=\frac14-d,\qquad b=\frac14,\qquad c=\frac14+d.$$ --- 2. **Use the G.P. condition on squares** Given that $a^2,b^2,c^2$ are in G.P., we have $$b^4=a^2c^2.$$ Since $a<b<c$ and $a+b+c=\frac34>0$, here all values come out positive, so we may also use $$b^2=ac.$$ Now substitute: $$\left(\frac14\right)^2=\left(\frac14-d\right)\left(\frac14+d\right).$$ Thus, $$\frac1{16}=\frac1{16}-d^2$$ which gives $$d^2=0.$$ That would force $a=b=c$, contradicting $a<b<c$. So let us carefully use the actual G.P. condition: $$\frac{b^2}{a^2}=\frac{c^2}{b^2}$$ $$b^4=a^2c^2$$ $$b^2=|ac|.$$ Because $a<b<c$ and one of them can be negative, we should allow $$ac=-b^2.$$ Now substitute: $$\left(\frac14-d\right)\left(\frac14+d\right)=-\left(\frac14\right)^2$$ $$\frac1{16}-d^2=-\frac1{16}$$ $$d^2=\frac18$$ $$d=\frac1{2\sqrt2}$$ (since $d>0$). Therefore, $$a=\frac14-d=\frac14-\frac1{2\sqrt2}.$$ --- 3. **Match with the options** This is exactly $$\boxed{\frac14-\frac1{2\sqrt2}}$$ which is **Option C**. --- 4. **Verification with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.