If the sixth term of an A.P. is zero, then \(\rm \dfrac{t_{18}}{t_{9}}\) is ?

Where tn denotes the nth term of AP.

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 = Sn =\(\rm \frac n 2\) [2a + (n − 1) × 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: Sixth term of an A.P. is zero

⇒ a6 = 0

⇒ a + (6 - 1) × d = 0

⇒ a + 5d = 0         

∴ a = -5d                     .... (1)

To Find: \(\rm \dfrac{t_{18}}{t_{9}}\)

\(\rm \Rightarrow \dfrac{t_{18}}{t_{9}} = \dfrac {a+17d}{a+8d}\)

\(\rm =\dfrac {-5d+17d}{-5d+8d}\\=\dfrac {12d} {3d}=4\)