Question

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

Answer 1

Answer:

Let a and d respectively be the first term and common difference of the AP.

Given a9 = 0

So, a + (9-1)d = 0

a+8d=0

a= -8d

Now, 29th term = a+28d

=-8d+28d

= 20d = 2 x 10d

= 2(-8d + 18d)

=2(a+18d)

= 2 x 19th term

Thus, the 29th term of the AP is twice the 19th term.

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Answer 2

Question

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

Answer : -

Given: 9th term of an A.P is 0 So, a9 = 0 We need to prove: a29 = 2a19

We know, an = a + (n – 1) d [where a is first term or a1 and d is common difference and n is any natural number] When n = 9: a9 = a + (9 – 1)d = a + 8d

According to question: a9 = 0 a + 8d = 0 a = -8d

When n = 19: a19

= a + (19 – 1)d = a + 18d

= -8d + 18d = 10d

When n = 29: a29

= a + (29 – 1)d

= a + 28d

= -8d + 28d

[Since, a = -8d] = 20d = 2×10d a29 = 2a19

[Since, a19 = 10d]

Hence Proved.