Find the common difference of an AP in which a18 – a14 = 32

Arithmetic Progression(A.P) is a sequence of numbers with a constant difference between consecutive terms. This difference is called the Common Difference of the AP. It is denoted as ‘d’. The first term of the AP is denoted as ‘a’. 

For Example, 1, 5, 9, 13 is an AP having first term, a = 1 and common difference, d = (5-1) = 4.

Formula for the nth term of an AP

Suppose we have an AP, whose

• First term = a,
• Common difference = d.

Therefore, its

• Second term, a₂ = First Term + d = a + d

Similarly, 

• Third term, a₃ = Second term + d = (a + d) + d = a + 2d

Observing the pattern, 

The formula for nth term:

a_n = a + (n – 1) × d

Formula for the Sum of an AP 

We have an AP having ‘n’ number of terms whose, 

First term = a, and 

Common difference = d.

For this AP, the formula for the Sum is, 

S_n = (n/2) × (a + l)

where l is the last term and 

l = (a + (n – 1) × d)

Another formula for sum is, 

S_n = (n/2) × (2a + (n-1) × d)

where we have substituted the value of l in terms of a and d.

Find the common difference of an AP in which a₁₈ – a₁₄ = 32

Answer:

Since, a_n = S_n – S_n-1

i.e. if we subtract the Sum of n terms of an AP from the Sum of (n-1) terms of the same AP, we get the nth term of the AP.

So, a₁₈ = S₁₈ – S₁₇   …..(1)

Similarly, a₁₄ = S₁₄ – S₁₃   ….(2)

Let the first term of the AP be a and the common difference be d.

Putting (1) and (2) into the given equation: a₁₈ – a₁₄ = 32, we get

(S₁₈ – S₁₇) – (S₁₄ – S₁₃) = 32

(18/2)(2 × a + (18-1) × d) – (17/2)(2 × a + (17-1) × d) – (14/2)(2 × a – (14-1) × d) + (13/2)(2 × a + (13-1) × d) = 32

Solving this equation we get d = 8.

Hence, the Common Difference of the AP in which a₁₈ – a₁₄ = 32 is 8.

Another Approach:

As we know that, a_n = a + (n-1)Xd

a₁₈ – a₁₄ = 32

a + 17 x d – (a + 13 x d) = 32

a + 17xd – a – 13xd = 32

4xd = 32

therefore, common difference: d = 8

Similar Questions

Question 1: In an AP, a₂=4, and a₄=10. Find the common difference(d) of the AP.

Answer: 

Since,

a₂=a+d=4   ….(1)

a₄=a+3 × d=10   ….(2)

Subtracting (2) from (1) i.e. a₄ – a₂

a + 3 x d – (a + d) = 10 – 4

2 × d = 6

Hence, d = 3.

Question 2: In an AP, a₆=4, and a₈=1. Find the common difference(d) of the AP.

Answer:

Since, 

a₆ = a+5 × d=4   ….(1)

a₈ = a+7 × d=1   ….(2)

Subtracting (2) from (1) i.e. a₈ – a₆

a + 7 x d – (a + 5 x d) = 4 -1

2 × d = -3

Hence, d = -1.5

Question 3: In an AP, a₃=14, and a₂=10. Find the first term(a) of the AP.

Answer:

Since, 

a₃ = a+2 × d=14   ….(1)

a₂ = a+d=10   ….(2)

Subtracting (1) from (2) i.e.  a₃ – a₂

a + 2 x d – (a + d) = 14 – 10

d = 4

Putting d=4 in equation (2), we get

a + d = 10 

a = 10-4 

a= 6

Hence, the first term is 6.

Question 4: In an AP, a₁₄=4, and a₇=11. Find the first term(a) of the AP.

Answer:

Since,

a₁₄=a+13 × d=4   ….(1)

a₇=a+6 × d=11   ….(2)

Subtracting (2) from (1) as,

7 × d = -7

d = -1

Putting d=-1 in equation (2), we get

a = 10 – d 

= 10 – (-1) 

= 10+1 

= 11

Question 5: Given the first and the last term of an AP with 10 terms are 10 and 40 respectively. Calculate the sum of the AP.

Answer:

Since, 

a = 10, l = 40, n = 10

Sum = (n/2) × (a+l)

= (10/2) × (10+40) 

= 5 × 50 

= 250

Question 6: Given the first term of an AP with 10 terms and the common difference of 10, is 10. Calculate the sum of the AP.

Answer:

Since, 

a = 10, d = 10, n = 10

Sum = (n/2) × (2 × a+(n-1) × d)

= (10/2) × (20+9 × 10)

= 5 × (20+90) 

= 5 × 110 

= 550.