Question

Refer the given sequence \[23, 21\frac{1}{2}, 20, ...\]

1. Find the general term of the given sequence.
2. Which term is the last positive term in the sequence.

Answer 1

(a) Given,

\[23, 21\frac{1}{2}, 20, ....\] is A.P.

\[a = 23\]

\[d = 21\frac{1}{2} - 23 = -1\frac{1}{2} = -\frac{3}{2}.\]

We know that,

\[a_n = a + (n - 1)d\]

⇒ \[a_n = 23 - \left(\frac{3}{2}\right)(n - 1)\]

⇒ \[a_n = 23 - \left(\frac{3n - 3}{2}\right)\]

⇒ \[a_n = \left(\frac{46 - 3n + 3}{2}\right)\]

⇒ \[a_n = \left(\frac{49 - 3n}{2}\right)\]

Hence, \[a_n = \left(\frac{49 - 3n}{2}\right).\]

(b) The last positive term occurs when:

⇒ \[a_n > 0\]

⇒ \[\frac{49 - 3n}{2} > 0\]

⇒ \[49 - 3n > 0\]

⇒ \[3n < 49\]

⇒ \[n < \frac{49}{3}\]

⇒ \[n < 16\frac{1}{3}\]

Thus, 16th term will be the last positive term of the sequence.

⇒ \[a_{16} = \left(\frac{49 - 3 \times 16}{2}\right)\]

= \[\frac{49 - 48}{2}\]

= \[\frac{1}{2}\]

= 0.5

Hence, the 16th term is the last positive term.