Question

If the 11^th  term of an AP is \(\frac {1}{13}\) and its 13^th  term is \(\frac {1}{11}\), then what will be the value of  143^th term?

(a) \(\frac {1}{143}\)

(b) 1

(c) 0

(d) \(\frac {23}{143}\)

This question was posed to me during a job interview.

This intriguing question originated from Nth Term of Arithmetic Progression in chapter Arithmetic Progressions of Mathematics – Class 10

Answer 1

Right choice is (b) 1

For explanation: Here 11^th  term = \(\frac {1}{13}\)

13^thterm = \(\frac {1}{11}\)

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

T11 = a + (n – 1)d = \(\frac {1}{13}\)

T11 = a + (11 – 1)d = \(\frac {1}{13}\)

T11 = a + 10d = \(\frac {1}{13}\)     (1)

T13 = a + (n – 1)d = \(\frac {1}{11}\)

T13 = a + (13 – 1)d = \(\frac {1}{11}\)

T13 = a + 12d = \(\frac {1}{11}\)     (2)

Subtracting (1) from (2)

We get,

a + 12d – (a + 10d) = \(\frac {1}{11} – \frac {1}{13}\)

2d = \(\frac {2}{143}\)

d = \(\frac {1}{143}\)

Now, substituting value of d in equation 1

We get,

T11 = a + 10(\(\frac {1}{143}\)) = \(\frac {1}{13}\)

a = \(\frac {1}{143}\)

The 143^th term = \(\frac {1}{143}\) + 142\(\frac {1}{143} = \frac {(1+142)}{143} \) = 1