Question

the 5th term of an arithematic Sequence is 20 and
the 8th term is 32
ما
Q What is the
common
difference of this sequence ?
Q find its 11th term​

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−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.

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