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# Question 7. If 9th term of an A.P. is zero, prove its 29th term is double the 19th term.

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One of the most important and exam oriented question from Chapter name- Arithmetic Progression
Class 10th
Chapter number- 9
Exercise :- 9.4
This type of question has been asked in previous years exams.
In this question we have been given that  9th term of an A.P. is zero, now we have to prove its 29th term is double the 19th term.

CBSE DHANPAT RAI publication
Understanding CBSE Mathematics
Question 7

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1. Solution:

Given that,

a9 = 0

As we know that, to find nth term in an A.P = a + (n – 1)d

therefore, a + (9 – 1)d = 0 ⇒ a + 8d = 0 ———-(i)

Now,

29th term is given by a29 = a + (29 – 1)d

=a29 = a + 28d

And, a29 = (a + 8d) + 20d (using (i))

= a29 = 20d ———-(ii)

Similarly, 19th term is given by a19 = a + (19 – 1)d

=a19 = a + 18d

And, a19 = (a + 8d) + 10d (using (i))

=a19 = 10d ———(iii)

On comparing (ii) and (iii), we observe that

a29 = 2(a19)

Hence, 29th term is double the 19th term.

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