In an A.P., the sixth term a_6=2 . If the product a_1 a_4 a_5 is the greatest, then the common difference of the A.P. is equal to

Sequences and Series PYQ — JEE Main 2024 · 29 Jan · Shift 1 · Q37 with Solution

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Correct answer: D

Let the first term be $a$ and common difference be $d$ . Then $a_n=a+(n-1)d$

Given: $a_6=a+5d=2$ So, $a=2-5d$

We have: $a_1=a=2-5d$ $a_4=a+3d=2-2d$ $a_5=a+4d=2-d$

Therefore, the product is $P=a_1a_4a_5=(2-5d)(2-2d)(2-d)$

We need the value of $d$ for which this product is greatest.

Let $x=2-d$ Then $2-2d=2x-2=2(x-1), \quad 2-5d=5x-8$ So $P=x(2x-2)(5x-8)=2x(x-1)(5x-8)$

This is a cubic expression, so instead it is easier to use the condition $a_6=2$ directly in a more elegant way.

Since $a_6=2$ ,

\quad a_4=2-2d, \quad a_1=2-5d$$ Thus $$P(d)=(2-5d)(2-2d)(2-d)$$ Expand: $$ (2-5d)(2-2d)=4-14d+10d^2 $$ So $$P(d)=(4-14d+10d^2)(2-d)$$ $$=8-32d+34d^2-10d^3$$ 5. **Differentiate to maximize** $$P'(d)=-32+68d-30d^2$$ Set $P'(d)=0$: $$-32+68d-30d^2=0$$ $$30d^2-68d+32=0$$ $$15d^2-34d+16=0$$ Solve: $$d=\frac{34\pm\sqrt{34^2-4\cdot 15\cdot 16}}{30}$$ $$=\frac{34\pm\sqrt{1156-960}}{30}$$ $$=\frac{34\pm 14}{30}$$ Hence, $$d=\frac{48}{30}=\frac{8}{5}$$ or $$d=\frac{20}{30}=\frac{2}{3}$$ 6. **Use second derivative test** $$P''(d)=68-60d$$ For $d=\frac{2}{3}$: $$P''\left(\frac{2}{3}\right)=68-40=28>0$$ So this gives a **minimum**. For $d=\frac{8}{5}$: $$P''\left(\frac{8}{5}\right)=68-96=-28<0$$ So this gives a **maximum**. Therefore, the product is greatest when $$\boxed{d=\frac{8}{5}}$$ 7. **Compare with stored answer** Stored correct answer: $D$ i.e. $\frac{8}{5}$. This matches our result.