Question

5th term of an arithmetic sequence is 17 and its 10th term is 32 .

a) What is its common difference ?

b) What is its first term ?

c) Find the position of 92 in this sequence ?​

Answer 1

GIVEN:

5th term of an Arithmetic sequence = 17

a + 4d = 17 ----(1)

10th term of an Arithmetic sequence = 32

a + 9d = 32 ----(2)

Subtract eq - (1) and (2)

a + 4d = 17

a + 9d = 32

----------------

-5d = -15

----------------

d = 15/5 = 3

Common Difference = 3

Substitute common difference (d) in eq - (1) to find a

=> a + 4d = 17

=> a + 4(3) = 17

=> a + 12 = 17

=> a = 17 - 12

=> a = 5

First Term = 5

an = a + (n - 1)d

92 = (5) + (n - 1)3

92 = 5 + 3n - 3

92 = 2 + 3n

=> 90 = 3n

=> n = 90/3

=> n = 30

Therefore, a = 5; d = 3 and 92 is 30th term.

Answer 2

Given:-

• 5th term of AP = 17
• 10th term of AP = 32

Solution:-

Now, 5th term = a + 4d

⇒ a + 4d = 17 - - - (Eq.1 )

Again, 10th term = a + 9d

⇒ a + 9d = 32 - - - (Eq.2 )

Now, subtracting (Eq.1) from (Eq.2)

⇒ a + 9d - a - 4d = 32 - 17

⇒ 5d = 15

• Dividing both terms by 5

⇒ d = 3

So, a) Common difference (d) = 3

$\rule{200}{1}$

Now,

Putting value of d in (Eq.1):-

⇒ a + 4(3) = 17

⇒ a + 12 = 17

⇒ a = 17 - 12

⇒ a = 5

So, b) First term (a) = 5

$\rule{200}{1}$

As we got a = 5 & d = 3

Formula for nth terms:-

☛ An = a + (n - 1)d

⇒ 92 = 5 + (n - 1)3

⇒ 92 = 5 + 3n - 3

⇒ 92 = 3n + 2

⇒ 92 - 2 = 3n

⇒ 3n = 90

• Dividing both terms by 3

⇒ n = 30

So,c)Position of 92 in this AP= 30th term