The sum of third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the sixteen terms of the AP.
Answers
A sequence that has a common difference between any pair of consecutive numbers is called an Arithmetic Progression.
nth 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 nth 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