Question

The first term of an (A.P) is -8. If the ratio of 7th and 9th term is5:8. Find the common difference of A.P
(2)sum of the first ten term

Answer 1

Step-by-step explanation:

a=-8

a7/a8=a+6d/a+8d =5/8

8(a+6d)=5(a+8d)

8a+48d=5a+40d

3a= -8d

3(-8)=-8d

d=3

S10= 10/2{2(-8)+(10-1)3}

S10=5{-16+9*3}

=5{-16+27}

=5(11) =55

Answer 2

Given that:

• The first term of an AP is - 8.
• The ratio of 7th and 9th term is 5 : 8.

To Find:

1. The common difference of AP.
2. Sum of the first ten term.

Let us assume:

• The common difference of AP be x.

We know that:

In an AP,

• aₙ = a + (n - 1)d
• Sₙ = n{2a + (n - 1)d}/2

Where,

• aₙ = nth term
• Sₙ = Sum of nth term
• a = First term
• n = Number of terms
• d = Common difference

Finding the common difference:

We have:

• a = - 8
• d = x

In 7th term,

• n = 7

In 9th term,

• n = 9

According to the question.

⟶ 7th term : 9th term = 5 : 8

⟶ 7th term / 9th term = 5 / 8

Cross multiplication.

⟶ 8 {7th term} = 5 {9th term}

Using formula.

⟶ 8 {a + (7 - 1)d} = 5 {a + (9 - 1)d}

Substituting the values.

⟶ 8 {- 8 + (7 - 1)x} = 5 {- 8 + (9 - 1)x}

⟶ 8 {- 8 + 6x} = 5 {- 8 + 8x}

⟶ - 64 + 48x = - 40 + 40x

⟶ 48x - 40x = - 40 + 64

⟶ 8x = 24

⟶ x = 24/8

⟶ x = 3

Hence,

The common difference of AP is 3.

Finding the sum of the first ten terms:

We have:

• n = 10
• a = - 8
• d = 3

⟶ Sₙ = n{2a + (n - 1)d}/2

Substituting the values.

⟶ S₁₀ = 10{2 × - 8+ (10 - 1)3}/2

⟶ S₁₀ = 10{- 16 + 9 × 3}/2

⟶ S₁₀ = 10{- 16 + 27}/2

⟶ S₁₀ = 10{11}/2

⟶ S₁₀ = 110/2

⟶ S₁₀ = 55

Hence,

• Sum of the first ten terms is 55.