If T₃ A. —4

B. 28

C. 32

D. 36

We have, T₃ = 8

And recall that, T_n = a + (n – 1)d

So we have, T₃ = a + (3 – 1)d

⇒ 8 = a + 2d ……(1)

We also have, T₇ = 24

And T₇ = a + (7 – 1)d

⇒ 24 = a + 6d …… (2)

From (1), we have that, a = 8 – 2d

So we put the value of a in (2)

⇒ 24 = 8 – 2d + 6d

⇒ 24 – 8 = 4d

⇒ 16 = 4d

⇒ d= 4

now, putting value of d in (1),

⇒ 8 = a + 2(4)

⇒ 8 = a + 8

⇒ a =0

now, T₁₀= a + (10 – 1)d

⇒ T₁₀= 0 + 9 × 4

⇒ T₁₀= 36

∴ correct option is (d).