Question

Which term of the ap 5,9,13,17,....,is 81

Answer 1

a=5
d=9-5=4
an=81
N=?
an=a+(n-1)d
81=5+(n-1)4
81-5=(n-1 )4
76= (n-1)4
76/4=n-1
19= n-1
19+1 =n
N= 20

Answer 2

Answer:

$20^{th}\: term \: in \: given \: A.P \: 81$

Step-by-step explanation:

Given A.P :5,9,13,17,...,81

$First\:term (a)=5\\common\: difference (d)=a_{2}-a_{1}\\=9-5\\=4\\last\:term (l)=81$

$Let\: n^{th}\: term(l) =81$

$a+(n-1)d=81$

$\implies 5+(n-1)4=81$

$\implies (n-1)4=81-5$

$\implies (n-1)4=76$

$\implies n-1=\frac{76}{4}$

$\implies n-1= 19$

$\implies n = 19+1$

$\implies n = 20$

Therefore,

$20^{th}\: term \: in \: given \: A.P \: 81$

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