1. If the 9th term
of the
A.P is zero
then show that
its 29th term is double of its 19th term.
Answers
Answered by
45
Step-by-step explanation:
Let the first term be 'a' and common difference be 'd'.
9th term = 0 → a + (9 - 1)d = 0
= > a + 8d = 0
= > a = - 8d
Now,
19th term is a + (19 - 1)d → a + 18d
From above, a = - 8d,
→ - 8d + 18d
→ 10d
29th term is a + (29 - 1)d → a + 28d
From above, a = - 8d,
→ - 8d + 28d
→ 20d
Result,
→ 20d = 2 * 10d
→ 29th term = 2 * 19th term
→ 29th term is double of its 19th term
Proved.
Answered by
73
Step-by-step explanation:
an = a + (n - 1) d
→ a9 = 0
→ a + (9 - 1)d = 0
→ a + 8d = 0
→ a = - 8d ................(1)
As per given condition,
a29 = 2 × a19 (which we have to prove)
a + 28d = 2 × (a + 18d)
-8d + 28d = 2a + 36d
20d = 2(-8d) + 36d
20d = -16d + 36d
20d = 20d
This proves that 29th term is double of its 19th term.
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