JEE Main 2020 — Sequences And Series Question with Solution

Let the A.P is
a - 2d, a - d, a, a + d, a + 2d
$$\because$$ sum = 25
$$\Rightarrow$$ 5a = 25 $$\Rightarrow$$ a = 5

Also given,

product (a² – 4d²) (a² – d²).a = 2520

$$\Rightarrow$$ (25 – 4d²) (25 –d²)5 = 2520

$$\Rightarrow$$ 4d⁴ – 121d² – 4d² + 121 = 0

$$\Rightarrow$$ (d² – 1) (4d² – 121) = 0

$$\Rightarrow$$ d = $$\pm$$1, d = $$\pm \frac{11}{2}$$

When d = $$\pm$$1 we can't get any fraction term like -$$\frac{1}{2}$$.

$$\therefore$$ d = $$\pm \frac{11}{2}$$

And when d = $$\frac{11}{2}$$

we get largest term = 5 + 2d = 5 + 11 = 16