If the 9th term of an A.P. is zero, then prove that 29th term is double of 19th term.
Solution 1
`t_n = a+ (n - 1)d`
9th term i.e n = 9
`:. t_9 = a+(9 -1)d`
= a + 8d
It is given that `t_9 = 0`
∴ a + 8d = 0 ....(i)
29th term i.e `t_29` where n = 29
`:. t_29 = a + (29-1)d`
`t_29 = a + 28d` ....(ii)
= (a + 8d) + 20d
= 0 + 20d ...by equ (i)
`:. t_29 = 20d` ....(ii)
`t_29 = a + (19 - 1)d`
`t_19` = a + 18d
= a + 8d + 10d
= 0 + 10d
`t_19 = 10d` .....(iiii)
by equation (ii) & (iii)
`t_29 = 2t_19`
Solution 2
In the given problem, the 9th term of an A.P. is zero.
Here, let us take the first term of the A.P as a and the common difference as d
So, as we know,
`a_n = a + (n - 1)d`
We get
`a_9 = a + (9 - 1)d`
0 = a + 8d
a = -8d .......(1)
Now, we need to prove that 29th term is double of 19th term. So, let us first find the two terms.
For 19th term (n = 19)
`a_19 = a + (19 - 1)d`
= -8d + 18d (using 1)
= 10d
For 29th term (n = 29)
`a_29 = a + (29 - 1)d`
`= - 8d + 28d`
= 20d
= 2 x 10d
`= 2 x a_19`
(Using 1)
Therefore for the given A.P. the 29th term is double of the 19th term.
Hence proved.