Question

The 6th term of an ap is 12 and the 8th is 22, find the 10 th term​

Answer 1

Answer :

• 10th term of the AP is 32.

Given :

• The 6th term of an AP is 12 and the 8th term is 22.

To Find :

• The 10th term.

Solution :

It is given that

• 6th term of the AP = 12
• 8th term of the AP = 22

Let

• 6th term of the AP => a + 5d = 12
• 8th term of the AP => a + 7d = 22

→ a + 5d = 12 ...i)

→ a + 7d = 22 ...ii)

Subtract eq ii) from i)

Now

→ a + 5d - (a + 7d) = 12 - 22

→ a + 5d - a - 7d = 12 - 22

→ 5d - 7d = - 10

→ - 2d = - 10

→ d = 10/2

→ d = 5

Put the value of d in eq i)

→ a + 5d = 12

→ a + 5(5) = 12

→ a + 25 = 12

→ a = 12 - 25

→ a = - 13

As we know that

• Tn = a + (n - 1)d

According to question :

→ Tn = - 13 + (10 - 1)5

→ Tn = - 13 + 9 × 5

→ Tn = - 13 + 45

→ Tn = 32

Hence, the 10th term of the AP is 32.

Answer 2

$\huge\boxed{\fcolorbox{purple}{ink}{Answer}}$

10th term of the AP is 32.

Given :

The 6th term of an AP is 12 and the 8th term is 22.

To Find :

The 10th term.

Solution :

It is given that

6th term of the AP = 12

8th term of the AP = 22

Let

6th term of the AP => a + 5d = 12

8th term of the AP => a + 7d = 22

→ a + 5d = 12 ...i)

→ a + 7d = 22 ...ii)

Subtract eq ii) from i)

Now

→ a + 5d - (a + 7d) = 12 - 22

→ a + 5d - a - 7d = 12 - 22

→ 5d - 7d = - 10

→ - 2d = - 10

→ d = 10/2

→ d = 5

Put the value of d in eq i)

→ a + 5d = 12

→ a + 5(5) = 12

→ a + 25 = 12

→ a = 12 - 25

→ a = - 13

As we know that

Tn = a + (n - 1)d

According to question :

→ Tn = - 13 + (10 - 1)5

→ Tn = - 13 + 9 × 5

→ Tn = - 13 + 45

→ Tn = 32

Hence, the 10th term of the AP is 32.