Determine the nth term of the AP whose 7th term is −1 and 16th term is 17.
Let a be the first term and d be the common difference of the AP. Then,
a7=−1
⇒a+(7−1)d=−1 [∵nth term is given by an=a+(n−1)d]
⇒a+6d=−1……(i)
Also, a16=17
⇒a+15d=17……(ii)
From eq. (i) and (ii), we get
−1−6d+15d=17 [∵a=−1−6d from eq.(i)]
⇒9d=17+1
⇒d=2
Putting d=2 in eq.(i), we get
a+6×2=−1
⇒a=−1−12=−13
∴an=a+(n−1)d=−13+(n−1)×2
=−13+2n−2=2n−15
Hence, the nth term of the AP is (2n−15).