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

**Solution:**

A sequence that has a common difference between any pair of consecutive numbers is called an Arithmetic Progression.

n^{th} term of an AP is aₙ = a + (n - 1)d

Here, a is the first term, d is the common difference and n is the number of terms.

Given:

- a₃ + a₇ = 6 ----- (1)
- a₃ × a₇ = 8 ----- (2)

We know that n^{th} term of AP is aₙ = a + (n - 1)d

Third term, a₃ = a + (3 - 1)d

a₃ = a + 2d ----- (3)

Seventh term,

a₇ = a + (7 - 1)d

a₇ = a + 6d ----- (4)

Using equation (3) and equation (4) in equation (1) to find the sum of the terms,

(a + 2d) + (a + 6d) = 6

2a + 8d = 6

a + 4d = 3

a = 3 - 4d ----- (5)

Using equation (3) and equation (4) in equation (2) to find the product of the terms,

(a + 2d ) × (a + 6d ) = 8

Substituting the value of a from equation (5) above,

(3 - 4d + 2d) × (3 - 4d + 6d) = 8

(3 - 2d) × (3 + 2d) = 8

(3)² - (2d)² = 8 [Since (a + b)(a - b) = a² - b² ]

9 - 4d² = 8

4d² = 1

d² = 1/4

d = ½, -½

Case 1: When d = ½

a = 3 - 4d

= 3 - 4 × ½

= 3 - 2

= 1

Sₙ = n/2 [2a + (n - 1) d]

S₁₆ = 16 / 2 [ 2 × 1 + (16 - 1) × ½ ]

= 8 × 19/2

= 76

Case 2: When d = - ½

a = 3 - 4d

= 3 - 4 × (- ½)

= 3 + 2

= 5

Sₙ = n/2 [2a + (n - 1) d]

S₁₆ = 16/2 [2 × 5 + (16 - 1) × (- ½)]

= 8 [10 - 15/2]

= 8 × 5/2

= 20

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 5

**Video Solution:**

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

Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.4 Question 2

**Summary:**

The sum of the third and the seventh terms of an A.P is 6 and their product is 8 then the sum of first 16 terms of the A.P. is equal to 20 or 76.

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