# 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.

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

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

Given:

- a
_{3}+ a_{7}= 6 ----- Equation(1) - a
_{3}× a_{7}= 8 ----- Equation(2)

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

Third term, a_{3} = a + (3 - 1)d

a_{3} = a + 2d ----- Equation(3)

Seventh term,

a_{7} = a + (7 - 1)d

a_{7} = a + 6d ----- Equation(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 ----- Equation (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 Equation (5) in above,

(3 - 4d + 2d) × (3 - 4d + 6d) = 8

(3 - 2d) × (3 + 2d) = 8

(3)² - (2d)² = 8

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

S_{n} = n/2 [2a + (n - 1) d]

S_{16} = 16/2 [2 × 5 + (16 - 1) × (- ½)]

= 8 [10 - 15/2]

= 8 × 5/2

= 20

Case 2: When d = - ½

a = 3 - 4d

= 3 - 4 × (- ½)

= 3 + 2

= 5

S_{n} = n/2 [2a + (n - 1) d]

S_{16} = 16 / 2 [ 2 × 1 + (16 - 1) × ½ ]

= 8 × 19/2

= 76

**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 :

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

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