Question

If the 9th term of an A.P is 0, prove that its 29th term is twice its 19th term

Answer 1

Let the first term, common difference and number of terms of an AP are a, d and n respectively.

Given that, 9th term of an AP, T9 = 0 [∵ nth term of an AP, Tn = a + (n-1)d]

⇒ a + (9-1)d = 0

⇒ a + 8d = 0 ⇒ a = -8d ...(i)

Now, its 19th term , T19 = a + (19-1)d

= - 8d + 18d [from Eq.(i)]

= 10d ...(ii)

and its 29th term, T29 = a+(29-1)d

= -8d + 28d [from Eq.(i)]

= 20d = 2 × T19

Hence, its 29th term is twice its 19th term

Answer 2

Answer:

Proved below.

Step-by-step explanation:

a9 = a1 + 8d = 0          where a1 = first term and d = common difference.

we need to prove that

a1 + 28d = 2(a1 + 18d

simplifying:-

a1  + 36d - 28d = 0

a1 + 8d = 0 which is what we are given.

Therefore  the proposition is true.