Question

Can zero be a term of an AP whose 11th term is 62 and 21st term is 32?​

Answer 1

Step-by-step explanation:

d= 32-62÷ 21- 11

= -30 ÷ 10

= -3

32 is not a multiple of 3. So, 0 is not a term in the sequence

Answer 2

Required Answer :

Zero cannot be a term of A.P.

Given :

• 11th term of A.P. = 62

• 21st term of A.P. = 32

To find :

• If zero can be a term of A.P.

Solution :

$\\ \longrightarrow \qquad \sf t_{11} = a + (n - 1)d = 62$

$\\ \longrightarrow \qquad \sf t_{11} = a + (11 - 1)d = 62$

$\\ \longrightarrow \qquad \sf t_{11} = a + (10)d = 62$

$\\ \longrightarrow \qquad \sf t_{11} = a + 10d = 62$

$\\ \longrightarrow \qquad \bf t_{11} = a + 10d = 62 \dots \dots \dots(1)$

$\\ \longrightarrow \qquad \sf t_{21} = a + (n - 1)d = 32$

$\\ \longrightarrow \qquad \sf t_{21} = a + (21 - 1)d = 32$

$\\ \longrightarrow \qquad \sf t_{21} = a + (20)d = 32$

$\\ \longrightarrow \qquad \bf t_{21} = a + 20d = 32 \dots \dots \dots (2)$

Solving (1) and (2) :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀a + 10d = 62

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀a + 20d = 32

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀-⠀-⠀⠀⠀⠀-

⠀⠀⠀⠀⠀⠀⠀⠀_______________

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀- 10d = 30

⠀⠀⠀⠀⠀⠀⠀⠀_______________

⇒ - 10d = 30

⇒ - d = 30/10

⇒ - d = 3

⇒ d = - 3

Substituting the value of d in equation (1) :

⇒ a + 10d = 62

⇒ a + 10(-3) = 62

⇒ a - 30 = 62

⇒ a = 62 + 30

⇒ a = 92

Therefore, the common difference of A.P. = - 3 and the first term of A.P. = 92

Finding the term of A.P. which is zero :

$\\ \longrightarrow \qquad \sf t_n = a + (n - 1)d$

$\\ \longrightarrow \qquad \sf 0 = 92 + (n - 1)(-3)$

$\\ \longrightarrow \qquad \sf - 92 = (n - 1)(-3)$

$\\ \longrightarrow \qquad \sf \dfrac{- 92}{-3} = (n - 1)$

$\\ \longrightarrow \qquad \sf \dfrac{- 92}{-3} = (n - 1)$

$\\ \longrightarrow \qquad \sf 30.67 = (n - 1)$

$\\ \longrightarrow \qquad \sf 30.67 + 1 = n$

$\\ \longrightarrow \qquad \sf 31.67 = n$

The nth term is not a whole number.

Therefore, zero cannot be a term of A.P. whose 11th term is 62 and 21st term is 32.