JEE Main 2019 — Sequences And Series Question with Solution

Let the three terms are a - d, a, a + d

Given a - d + a + a + d = 33

$$\Rightarrow$$ 3a = 33

$$\Rightarrow$$ a = 11

Also given,

(a - d)a(a + d) = 1155

$$\Rightarrow$$ (a² - d²)a = 1155

$$\Rightarrow$$ (11² - d²)11 = 1155

$$\Rightarrow$$ (11² - d²) = 105

$$\Rightarrow$$ d = $$\pm$$ 4

When d = 4 and a = 11 then series is

7, 11, 15, ....

$$\therefore$$ T₁₁ = a + 10d = 7 + 10$$\times$$ 4 = 47

When d = -4 and a = 11 then series is

15, 11, 7, ....

$$\therefore$$ T₁₁ = a + 10d = 15 + 10$$\times$$ -4 = -25