If T₄ = 7 and T₇ = 4, then T₁₀ = …..

Given T₄ = 7

And we know that, T_n = a + (n – 1)d

So we have T₄ = a + (4 – 1)d

⇒ 7 = a + 3d …… (1)

Then, T₇ = 4

⇒ T₇ = a + (7 – 1)d

⇒ 4 = a + 6d …….(2)

Now, (2) – (1) gives :

⇒ – 3 = 3d

⇒ d = – 1

so, now, putting value of d in (1),

⇒ 7 = a + 3d

⇒ 7 = a + 3( – 1)

⇒ 7 = a – 3

⇒ a = 10

so now T₁₀ = a + (10 – 1)d

⇒ T₁₀ = a + 9d

= 10 + 9( – 1)

= 10 – 9

= 1

∴ T₁₀ = 1

∴ correct option is (d)