IF 9th TERM OF AN A.P. IS ZERO THEN PROVE THAT ITS 29th TERM IS DOUBLETHE 19th TERM
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let 'a' and 'd' be the first term and common difference of an A.P
we have,
a9 = 0.
=> a = -8d --- (1)
now,
a29 = a + 28d
=> a29 = -8d + 28d [from (1)]
=> a29 = 20d --- (2)
and a19 = a+18d
=> a19 = -8d + 18d [from (1)]
=> a19 = 10d --- (3)
from (2) and (3)
we have
20d = 2×10d
=> a29 = 2×a19, proved
we have,
a9 = 0.
=> a = -8d --- (1)
now,
a29 = a + 28d
=> a29 = -8d + 28d [from (1)]
=> a29 = 20d --- (2)
and a19 = a+18d
=> a19 = -8d + 18d [from (1)]
=> a19 = 10d --- (3)
from (2) and (3)
we have
20d = 2×10d
=> a29 = 2×a19, proved
Answered by
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Question:-
➡ If the 9th term of an A.P. is zero then prove that, 29th term is twice the 19th term.
Proof:-
Let us assume that,
➡ First term of the A.P. = a and,
➡ Common Difference = d
Now,
Nth term of an A.P. = a + (n -1)d
So,
9th term = a + (9 - 1)d
= a + 8d
Now, it's given that, 9th term of the A.P. is zero.
➡ a + 8d = 0 .....(i)
Now,
29th term = a + (29 - 1)d
= a + 28d
19th term = a + (19 - 1)d
= a + 18d
Now,
29th term - 2 × 19th term
= a + 28d - 2 × (a + 18d)
= a + 28d - 2a - 36d
= -a - 8d
= -1(a + 8d)
= -1 × 0
= 0
Hence,
29th term - 2 × 19th term = 0
➡ 29th term = 2 × 19th term. (Hence Proved)
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