Question

Q8 The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first

sixteen terms of the AP.​

Answer 1

Answer:

If d = 1/2, sum of 16 terms is 76.

If d = -1/2, sum of 16 terms is 20.

Step-by-step explanation:

Given:

• Sum of the third and seventh term of an A.P is 6
• Product of the the third and seventh term of the A.P is 8

To Find:

• Sum of the first sixteen terms of the A.P

Concept:

Here we have to make a linear equation in two variables with both the given cases. Solving it we get the value of the variables. Substituting the vallue for finding the sum of n terms we get the value of sum of sixteen terms of the A.P.

Solution:

First we have to find the third and seventh term of the A.P

The third term of an A.P is given by,

Third term = a₁ + 2d

Seventh term of the A.P is given by,

Seventh term = a₁ + 6d

By first case given,

a₁ + 2d + a₁ + 6d = 6

2a1 + 8d = 6

Divide the whole equation by 2

a1 + 4d = 3

a₁ = 3 - 4d-------(1)

Now by second case given,

(a₁ + 2d) × (a₁ + 6d) = 8

Substitute the value of a₁ from equation 1

(3 - 4d + 2d) × (3 - 4d + 6d) = 8

(3 - 2d) × (3 + 2d) = 8

By using the identity,

(a + b) × (a - b) = a² - b²,

9 - 4d² = 8

-4d² = -1

d² = 1/4

d = ± 1/2

Case 1:

If d = 1/2

Substitute the value of d in equation 1,

a₁ = 3 - 4 × 1/2

a₁ = 3 - 2

a₁ = 1

Now sum of n terms of an A.P is given by,

$\sf{S_n=\dfrac{n}{2} (2a_1+(n-1)\times d)}$

Substitute the data,

$\sf{S_{16}=\dfrac{16}{2}(2\times 1+(16-1)\times \frac{1}{2} )}$

S₁₆ = 8 (2 + 15/2)

S₁₆ = 8 (2 + 7.5)

S₁₆ = 8 × 9.5

S₁₆ = 76

Hence sum of 16 terms when d = 1/2 is 76

$\boxed{\bold{If\:d=\dfrac{1}{2},S_{16}=76}}$

Case 2:

If d = -1/2

Substitute the value of d in equation 1

a₁ = 3 - 4 × -1/2

a₁ = 3 + 2

a₁ = 5

Hence first term of the A.P is 5

Substituting the value in the formula for finding sum of n terms,

$\sf{S_{16}=\dfrac{16}{2}(2\times 5+(16-1)\times -\frac{1}{2} )}$

S₁₆ = 8 (10 + 15 × -1/2)

S₁₆ = 8 ( 10 - 7.5)

S₁₆ = 8 × 2.5

S₁₆ = 20

Hence the sum of 16 terms when d = -1/2 is 20

$\boxed{\bold{If\:d=-\dfrac{1}{2},S_{16}=20}}$