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:
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.
