Question

Question 9
If the $9^{th}$ term of an AP is zero, then prove that its ${29}^{th}$ term is twice its ${19}^{th}$ term.

Answer 1

Let the first term, common difference and number of terms of an AP are a, d and n respectively.
Given that,
$9^{th}$ term of the AP, $T_9$ = 0
[$\because$ nth term of an Ap, $T_n$ = a + (n - 1)d]
a + (9 - 1)d = 0
$\Rightarrow$ a + 8d = 0 . . . .(i)
19th term, $T_{19}$ = a + (19 - 1)d
= - 8d + 18d [From eq. (i)]
= 10d . . . . (ii)
29th term, $T_{29}$ = a + (29-1)d
= - 8d + 28d [from Eq.(ii)]
= 20d = 2$\times$(10d)
$\Rightarrow T_{29}=2\times T_{19}$