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:

First 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

Answer 2

Answer:

Step-by-step explanation:

Given AP is 1,5,9,13,....

$\tt{1st\,\,term,\,a=1\,\,\,\,\,\,\,common\,\,\,\,difference,\,\,d=4}$

Let $n^{th}$, term be 101, i.e.,

$\sf{a_{n}=101}$

$\sf{\implies\,a+(n-1)d=101}$

$\sf{\implies\,1+(n-1)4=101}$

$\sf{\implies\,(n-1)4=101-1}$

$\sf{\implies\,(n-1)4=100}$

$\sf{\implies\,(n-1)=\dfrac{100}{4}}$

$\sf{\implies\,n-1=25}$

$\sf{\implies\,n=25+1}$

$\sf{\implies\,n=26}$

$\boxed{\rm{\green{Hence,101\,\,is\,\,the\,\,26^{th}\,\,term\,\,of\,\,the\,\,given\,\,AP}}}$