Question

Which term of A.P. 1,5,9,13.... is 101 ?? ​

Answer 1

Answer:

★Hence,the $26^{th}$ term of the given AP is 101

Step-by-step explanation:

★Given:

$1^{st}$ term = a = 1

Difference = d = 5 - 1 = 4

a(n) = 101

★To find:

$n^{th}$ term

★Solution:

⇒ aₙ = a + (n-1)d

⇒ 101 = 1 + (n-1)4

⇒101 = 1 + 4n-4

⇒101 = 4n-3

⇒4n = 104

⇒ n = $\frac{104}{4}$

⇒ n = 26

★Hence,the $26^{th}$ term of the given AP is 101