Question

If the 9^th term of an A.P. is zero, then prove that 29^th term is double of 19^th term.

Answer 1

t_n = a + (n – 1)d

9^th term i.e., n = 9

∴ t₉ = a + (9 – 1)d

= a + 8d

It is given that t₉ = 0

∴ a + 8d = 0 ....(i)

29^th term i.e t₂₉ where n = 29

∴ t₂₉ = a + (29 – 1)d

t₂₉ = a + 28d ....(ii)

= (a + 8d) + 20d

= 0 + 20d      ......By equation (i)

∴ t₂₉ = 20d ....(ii)

t₁₉ = a + (19 – 1)d`

t₁₉ = a + 18d

= a + 8d + 10d

= 0 + 10d

t₁₉ = 10d .....(iii)

By equation (ii) and (iii)

t₂₉ = 2t₁₉

Answer 2

In the given problem, the 9^th term of an A.P. is zero.

Here, let us take the first term of the A.P as a and the common difference as d

So, as we know,

a_n = a + (n – 1)d

We get

a₉ = a + (9 – 1)d

0 = a + 8d

a = – 8d .......(1)

Now, we need to prove that 29^th term is double of 19^th term. So, let us first find the two terms.

For 19^th term (n = 19)

a₁₉ = a + (19 – 1)d

= – 8d  + 18d    .....(Using 1)

= 10d

For 29^th term (n = 29)

a₂₉ = a + (29 – 1)d

= – 8d + 28d

= 20d

= 2 × 10d

= 2 × a₁₉  ......(Using 2)

Therefore for the given A.P. the 29^th term is double the 19^th term.

Hence proved.