(i) \(7, 13, 19, ..…, 205\)
For this A.P., the first term \(a = 7\) and the common difference \(d = a_2 − a_1 = 13 − 7 = 6\).
Let there be \(n\) terms in this A.P. The last term is \(a_n = 205\).
Using the formula \(a_n = a + (n − 1) d\), we have:
\(205 = 7 + (n − 1) 6\)
\(198 = (n − 1) 6\)
\(33 = (n − 1) \)
\(n = 34\)
Therefore, this given series has 34 terms.
(ii) \(18,15\frac 12 ,13, ..…,−47\)
For this A.P., the first term \(a = 18\).
The common difference \(d = a_2-a_1 = 15 \frac 12 -18\).
\(d = \frac {31}{2} - 18\)
\(d= \frac {31-36}{2}= -\frac 52\)
Let there be \(n\) terms in this A.P. The last term is \(a_n = −47\).
Using the formula \(a_n = a +(n-1)d\), we have:
\(-47 = 18 + (n-1)(-\frac 52)\)
\(-47-18 = (n-1)(-\frac 52)\)
\(-65 = (n-1)(-\frac 52)\)
\(\frac {-65 \times 2}{-5} = n-1\)
\(n-1 = \frac {-130}{-5}\)
\(n-1 = 26\)
\(n = 27\)
Therefore, this given A.P. has 27 terms.