Correct Answer - Option 2 : 5

__Concept:__

Let us consider sequence a1, a2, a3 …. an is an A.P.

- Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1

- nth term of the A.P. is given by an = a + (n – 1) d
- Sum of the first n terms = \(S_n= \frac{n}{2}[2a+(n-1)\times d]\) = \(\rm\frac{n}{2}[a+l]\)

Where, a = First term, d = Common difference, n = number of terms, an = nth term and l = Last term

__Calculation:__

Let the first term of AP be 'a' and the common difference be 'd'

Given: Ninth term of an A.P. is zero

⇒ a_{7} = 0

⇒ a + (7 - 1) × d = 0

⇒ a + 6d = 0

∴ a = -6d ----(1)

To find: \(\frac{{{t_{37}}}}{{{t_{13}}}}\)

⇒ \(\frac{{{t_{37}}}}{{{t_{13}}}} = \frac{{a + \;36d}}{{a + 12d}}\)

⇒ \(\frac{{{t_{37}}}}{{{t_{13}}}} = \frac{{ - 6d\; + \;36d}}{{ - 6d\; + \;12d}}\)

⇒ \(\frac{{{t_{37}}}}{{{t_{13}}}} = \frac{{\;30d}}{{\;6d}}\)

**∴ \(\frac{{{t_{37}}}}{{{t_{13}}}}\) = 5**