Question

1) ​​Is 101 a term of the arithmetic sequence 13,24,35.... ? what about 1001?
Please answer me..​

Answer 1

Given:-

• AP = 13,24,35...$\sf a_n$

• First term (a) = 13

• Common diffrence (d) = $\sf a_2 - a_1 = 24 - 13 = 11$

To Show:-

1. Is 101 is a term of the given AP
2. Is 1001 is a term of the given AP

Solution:-

★ Let 101 be $\sf a_n$ term of AP

$\star\;\;\sf {\underline{ a_n = a + (n - 1)d}}$

$\dashrightarrow$ 101 = 13 + (n - 1)11

$\dashrightarrow$ 101 = 13 + 11n - 11

$\dashrightarrow$ 101 = 11n + 2

$\dashrightarrow$ 101 - 2 = 11n

$\dashrightarrow$ 99 = 11n

$\dashrightarrow\sf n = \cancel{ \dfrac{99}{11}}$

$\dashrightarrow\sf \bf{\red{n = 9}}$

Since, n can be 9,

therefore, 101 is one of the term of the given AP

$\rule{200}3$

★ Let 1001 be $\sf a_n$ term of AP

$\star\;\;\sf {\underline{ a_n = a + (n - 1)d}}$

$\dashrightarrow$ 1001 = 13 + (n - 1)11

$\dashrightarrow$ 1001 = 13 + 11n - 11

$\dashrightarrow$ 1001 = 11n + 2

$\dashrightarrow$ 1001 - 2 = 11n

$\dashrightarrow$ 999 = 11n

$\dashrightarrow\sf n = \dfrac{999}{11}$

$\dashrightarrow\sf \bf{\red{n = 90.81...}}$

Since, n can't be a fraction,

therefore, 1001 is not a term of the given AP

$\rule{200}3$