Question

If 5th term of an AP is zero,prove that 23rd term is three times its 11th term

Answer 1

t5=0
a+4d=0
a=-4d.................................(1)
To prove:t23=3(t11)
Proof: t23=a+22d
=-4d+22d...........[from(1)]
t23 =18d
Now,
t11=a+10d
=-4d+10d
t11 =6d
as 18d=3(6d)
so,
t23=3(t11)
OK!!!

Answer 2

$a_{23} =3\times a_{11}$, proved.

Step-by-step explanation:

Let first term = a and common difference = d

Given,

$a_{5} =0$

Prove that, $a_{23} =3\times a_{11}$

∴ $a_{5} =a+(5-1)d=a+4d$

⇒ $a+4d=0$

⇒$a=-4d$     .....(1)

$a_{23} =a+(23-1)d=a+22d$

Using (1), we get

$a_{23} =-4d+22d=18d$

Also,

$a_{11} =a+(11-1)d=a+10d$

Using (1), we get

$a_{11} =-4d+10d=6d$

$a_{23} =3\times a_{11}$=18d, proved.