Question

The 5th term of an arithmetic sequence is 20 and the 8th term is 32 . (a) What is the common difference of this sequence? (b) Find its 11th term​

Answer 1

Solution:

Let us assume that:

• aₙ = nth term of the AP.
• d = common difference of the AP.
• a = First term of the AP.

Therefore, we can write:

→ aₙ = a + (n - 1)d

Given That:

→ a₅ = 20

→ a₈ = 32

Or, we can say:

→ a + 4d = 20 — (i)

→ a + 8d = 32 — (ii)

Subtracting (ii) from (i), we get:

→ -4d = -12

→ d = 3

★ So, the common difference of the AP is 3.

Substituting the value of d in (i), we get:

→ a + 4 × 3 = 20

→ a + 12 = 20

→ a = 8

★ So, the first term is 8 and common difference is 3.

Therefore, the 11th term will be:

→ a₁₁ = a + 10d = 8 + 30 = 38

★ Therefore, the 11th term of the AP is 38.

Answer:

• The common difference of the AP is 3.
• 11th term of the AP is 38.

To Know More:

Let us assume that:

→ First term = a

→ Common Difference = d

→ Nth term = aₙ

→ Sum of first n terms = Sₙ

Then:

→ aₙ = a + (n - 1) × d — (i)

→ aₘ - aₙ = (m - n) × d — (ii)

→ Sₙ = n[2a + (n - 1)d]/2 — (iii) or,

→ Sₙ = n(a + l)/2 where l is the last term of the AP. — (iv)

Answer 2

GIVEN :-

• 5th term of A.P is 20.
• 8th term of A.P is 32.

TO FIND :-

• 11th term.
• Common difference (d).

TO KNOW :-

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

Here ,

• a{n} → 'n'th term.
• a → 1st term.
• n → Number of terms.
• d → Common difference.

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)

Also,

• ♦ 8th term is 32.

We have,

• a{n} = 32
• n = 8

Putting values ,

• → 32 = a + (8-1)d
• → 32 = a + 7d ---------(2)

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.

Putting d = 4 in equation (1) ,

→ 20 = a + 4d

→ 20 = a + 4(4)

→ 20 = a + 16

→ a = 20 - 16

→ a = 4

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.