If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.
Answers
Step-by-step explanation:
9th term=a+(9-1)d
=a+8d
a+8d=0
=>a=-8d .....(i)
29th term= a+(29-1)d
= a+28d
=-8d+28d ......(using (i))
=20d
19th term= a+(19-1)d
= a+18d
= -8d+18d ......(using (i))
= 10d
19th term=10d
19th term *2=10d*2
=20d
=29th term (PROVED)
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)