If the 9 th term of an A.P. is zero,then prove that its 29 th term is twice its 19 th term.
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The nth term of an AP is given by a + (n-1) d
Hence, the 9th term is 0 and can be expressed as a + (9-1) d = a + 8d = 0.
The 19th term is a + (19-1) d = a + 18d = a + 8d + 10d.
Since a + 8d = 0, the 19th term = 10d.
Similarly, the 29th term = a+(29-1)d = a + 28d = a + 8d + 20d = 20d, since a+8d = 0.
Thus, 29th term = 20d = 2 x 10d = 2 x 19th term.
Hence, proved for this AP, the 29th term = twice 19th term.
Hence, the 9th term is 0 and can be expressed as a + (9-1) d = a + 8d = 0.
The 19th term is a + (19-1) d = a + 18d = a + 8d + 10d.
Since a + 8d = 0, the 19th term = 10d.
Similarly, the 29th term = a+(29-1)d = a + 28d = a + 8d + 20d = 20d, since a+8d = 0.
Thus, 29th term = 20d = 2 x 10d = 2 x 19th term.
Hence, proved for this AP, the 29th term = twice 19th term.
Answered by
6
Answer:
Step-by-step explanation:
nth term= a+(n-1)d
hence,9th term= a+(9-1)d
=a+8d
a= -8d
19th term= a+18d
= -8d+18d
= 10d
29th term=a+28d
= -8d+28d
= 20d
therefore,
19th term's double is 20th term i;e
a29= 2(a19)
hence proved..
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