Question

Which term of the following A.P is 560?

2, 11, 20, 29,......

Answer 1

Here, a = t₁ = 2, t₂ = 11, t₃ = 20, t₄ = 29, ...

d = t₂ – t₁ = 11 – 2 = 9.

Let the nth term of the given A.P. be 560.

Then t_n = 560.

t_n = a + (n – 1)d   ...(Formula)

∴ 560 = 2 + (n – 1) × 9   ...(Substituting the values)

∴ 2 + (n – 1) × 9 = 560

∴ (n – 1) × 9 = 560 – 2

∴ (n – 1) × 9 = 558

∴ `n - 1 = 558/9`

∴ n – 1 = 62

∴ n = 62 + 1

∴ n = 63

The 63rd term of the given A.P. is 560.