Question

Find the 19th term of the following sequence: tn = n2 , where n is even and tn = n2 - 1 , where n is odd

Answer 1

Given that

n = 19
19 is an odd number

Now
t19 = (19^2- 1)
= 360 - 1
= 360

Answer 2

Answer:

$\red { 19^{th} \: term \: in \: sequence }\green {=360}$

Step-by-step explanation:

$Given \: sequence \:\red { t_{n} = n^{2}\: where \: n \: is \: even }$

$and \: \red { t_{n} = n^{2}-1\: where \: n \: is \: odd }$

$If \: n = 19 \: (Odd)$

$t_{n} = n^{2} - 1$

$\implies t_{19} = 19^{2} - 1\\=361 - 1\\= 360$

Therefore.,

$\red { 19^{th} \: term \: in \: sequence }\green {=360}$

•••♪