Question

If the seventh term of an A.P. is zero, then \(\frac{{{t_{37}}}}{{{t_{13}}}}\) is ? 

 tn denotes the nth term of AP.

1. 3
2. 5
3. 1
4. 2

Answer 1

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

⇒ a₇ = 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