Question

check wheather 101 is term of ap 12,15, 18​

Answer 1

Solution -

We have,

• A.P = 12, 15, 18, ...

⠀

Here,

• First term (a) = 12
• Common difference (d) = 3

⠀

Now, we have to check whether 101 is a term of the given A.P. or not. For this, we let

• aₙ = 101

⠀

We know that,

⠀ ⠀⠀⠀⠀★ aₙ = a + (n - 1)d

⠀

➝ aₙ = 101

➝ a + (n - 1)d = 101

➝ 12 + (n - 1) × 3 = 101

➝ (n - 1)3 = 101 - 12

➝ (n - 1)3 = 99

➝ n - 1 = 99/3

➝ n - 1 = 33

➝ n = 33 + 1

➝ n = 34

⠀

Therefore, 34th term of the given A.P. is 101.

Hence, 101 is a term of A.P.

Answer 2

$\bold{ We \: have \: :- }$

• A.P = 12,15,18____

$\bold{ Here }$

• First term ( a) = 12
• Common difference ( d) = 3

Now, we have to check whether 101 is a term of the given A.P or not. For this, we let :-

${ A_n = 101 }$

We, know that :-

• ${ A_n = 4+(n-1)d }$

${A_n = 101 }$

$\\ a + (n - 1)d = 101 \\ 12 + (n - 1) \times 3 = 101 \\ (n - 1)3 = 101 - 12 \\ (n - 1)3 = 99 \\ n - 1 = \frac{99}{3} \\ n - 1 = 33 \\ n = 33 + 1 \\ n = 34$

therefore, 34th term of given A.P is 101 .

Hence, 101 is a term of A.P