in an A.P if the sum of the third term and seventh term is 6 and their product is 8, then find the sum of 16terms
Answers
Solution :-
Third term = a₃ = a + (3 - 1)d = a + 2d
Seventh term = a₇ = a + (7 - 1)d = a + 6d
Sum of third term and seventh term = 6
⇒ a + 2d + a + 6d = 6
⇒ 2a + 8d = 6
⇒ a + 4d = 3 -- eq(1)
Product of third term and seventh term = 8
⇒ (a + 2d)(a + 6d) = 8
It can be written as
⇒ (a + 4d - 2d) (a + 4d + 2d) = 8
⇒ (3 - 2d) (3 + 2d) = 8
[ From eq(1) ]
⇒ 3² - (2d)² = 8
⇒ 9 - 4d² = 8
⇒ 9 - 8 = 4d²
⇒ 1 = 4d²
⇒ 1/4 = d²
⇒ d = 1/2 or - 1/2
Substituting d = 1/2 in eq(1)
⇒ a + 4d = 3
⇒ a + 4(1/2) = 3
⇒ a + 2 = 3
⇒ a = 1
By using Sum of n terms of an AP formula
Sₙ = n/2(2a + (n - 1)d)
⇒ S₁₆ = 16/2 [2(1) + (16 - 1)(1/2)]
⇒ S₁₆ = 8(2 + 15/2)
⇒ S₁₆ = 8[ (4 + 15)/2 ]
⇒ S₁₆ = 8(19/2)
⇒ S₁₆ = 76
Therefore sum of 16 terms is 76.
Answer: 76
Step by step explanation:
Given:
Sum of third and seventh is 6 and their product is 8.
Third term = a₃ = a + (3 - 1)d = a + 2d
and seventh term = a₇ = a + (7 - 1)d = a + 6d
Now, Sum of third term and seventh term is 6 [Given]
=> a + 2d + a + 6d = 6
=> 2a + 8d = 6
=> a + 4d = 3 __(1)
Product of third term and seventh term = 8 [Given]
=> (a + 2d)(a + 6d) = 8
From (1),
=> (a + 4d - 2d) (a + 4d + 2d) = 8
=> (3 - 2d) (3 + 2d) = 8
=> 3² - (2d)² = 8
=> 9 - 4d² = 8
=> -4d² = 8-9
=> d² = 1/4
.°. d = 1/2
Substituting d = 1/2 in (1),
=> a + 4d = 3
=> a + 4(1/2) = 3
=> a + 2 = 3
=> a = 3 - 2
=> a = 1
We've to find S₁₆,
By using the formula:
Sₙ = n/2[2a + (n - 1)d]
=> S₁₆ = 16/2 [2×1 + (16 - 1)×(1/2)]
=> S₁₆ = 8[2 + 15/2]
=> S₁₆ = 8[(4/2 + 15/2)]
=> S₁₆ = 8(19/2)
=> S₁₆ = 4 × 19
=> S₁₆ = 76
Thus, Answer => 76