Question

find the common difference and first term of a AP in which a18-a14=32 .​

Answer 1

Answer:

let common difference be d

a18= a+17d

a14=a+13d

a+17d-(a+13d)=32

a+17d-a-13d=32

17d-13d=32

4d=32

d=32/4

d=8

common difference is 8

Answer 2

Answer:

The first term and common difference of an AP are 0 & 8 respectively.

Step-by-step-explanation:

We have given that,

The difference between the 18ᵗʰ and 14ᵗʰ term of an AP is 32.

We have to find the first term and common difference of the AP.

Now, we know that,

tₙ = a + ( n - 1 ) * d - - - [ Formula ]

Now,

t₁₈ - t₁₄ = 32 - - - [ Given ]

⇒ [ a + ( 18 - 1 ) * d ] - [ a + ( 14 - 1 ) * d ] = 32

⇒ ( a + 17 * d ) - ( a + 13 * d ) = 32

⇒ a + 17d - ( a + 13d ) = 32

⇒ a + 17d - a - 13d = 32

⇒ a - a + 17d - 13d = 32

⇒ 0 + 4d = 32

⇒ 4d = 32

⇒ d = 32 ÷ 4

⇒ d = 8

Now,

t₁₈ = a + ( 18 - 1 ) * d

⇒ t₁₈ = a + 17 * 8

⇒ t₁₈ = a + 136

Now,

t₁₄ = a + ( 14 - 1 ) * d

⇒ t₁₄ = a + 13 * 8

⇒ t₁₄ = a + 104

Now,

t₁₈ - t₁₄ = 32

⇒ a + 136 - ( a + 104 ) = 32

⇒ a + 136 - a - 104 = 32

⇒ a - a + 32 = 32

⇒ a - a = 32 - 32

⇒ 0 = 0

∴ a = 0

∴ The first term and common difference of an AP are 0 & 8 respectively.