Question

The 5 th term of an arithmetic sequence is 20 and 8th term is 32.

b. Is 145 a term of this sequence justify?

c. write the algebraic from of this sequence? ​

Answer 1

Answer:

GIVEN :-

5th term of A.P is 20.

8th term of A.P is 32.

\begin{gathered} \\ \end{gathered}

TO FIND :-

11th term.

Common difference (d).

\begin{gathered} \\ \end{gathered}

TO KNOW :-

\begin{gathered} \\ \bigstar \boxed{ \sf \: a_{n} = a + (n - 1)d} \\ \end{gathered}

★

a

n

=a+(n−1)d

Here ,

a{n} → 'n'th term.

a → 1st term.

n → Number of terms.

d → Common difference.

\begin{gathered} \\ \end{gathered}

SOLUTION :-

♦ 5th term of A.P is 20.

We have ,

n = 5

a{5} = 20

Putting values ,

20 = a + (5 - 1)d

20 = a + 4d --------(1)

\begin{gathered} \\ \end{gathered}

Also,

♦ 8th term is 32.

We have,

a{n} = 32

n = 8

Putting values ,

→ 32 = a + (8-1)d

→ 32 = a + 7d ---------(2)

\begin{gathered} \\ \end{gathered}

Subtracting equation (2) by equation (1) ,

→ 20 - 32 = a + 4d - (a + 7d)

→ -12 = a + 4d - a - 7d

→ -12 = -3d

→ d = -12/-3

→ d = 4

Hence , Common difference is 4.

\begin{gathered} \\ \end{gathered}

Putting d = 4 in equation (1) ,

→ 20 = a + 4d

→ 20 = a + 4(4)

→ 20 = a + 16

→ a = 20 - 16

→ a = 4

\begin{gathered} \\ \end{gathered}

Now , we will find 11th term of A.P.

→ a{11} = a + (11-1)d

Putting values of a and d,

→ a{11} = 4 + 10(4)

→ a{11} = 4 + 40

→ a{11} = 44

Hence , 11th term of the A.P is 44.