the sum of the 3rd and the 7th terms of an AP is 6 and their product is 8. Find the sum of the first 20 terms of the AP.
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Answers
Answer:
115 (or) 5
Step-by-step explanation:
nth term of an AP a(n) = a + (n - 1) * d.
(i) Sum of 3rd and 7th term of an AP is 6:
a₃ + a₇ = 6
(a + 2d) + (a + 6d) = 6
a + 2d + a + 6d = 6
2a + 8d = 6
a + 4d = 3
a = 3 - 4d.
(ii) Product of 3rd and 7th terms of an AP is 8:
a₃ * a₇ = 8
(a + 2d) * (a + 6d) = 8
(3 - 4d + 2d) * (3 - 4d + 6d) = 8
(3 - 2d) * (3 + 2d) = 8
9 - 4d² = 8
-4d² = -1
d² = 1/4
d = ± 1/2.
When d = 1/2:
a + 4d = 3
a + 4(1/2) = 3
a + 2 = 3
a = 1.
When d = -1/2:
a + 4d = 3
a + 4(-1/2) = 3
a - 2 = -3
a = 3 + 2
a = 5
Hence, a = 1, d = 1/2 (or) a = 5, d = -1/2.
Sum of first 20 terms:
When a = 1, d = 1/2:
S₂₀ = (20/2)[2(1) + (20 - 1) * 1/2]
= 10[2 + 19/2]
= 10[23/2]
= 115.
When a = 5, d = -1/2:
S₂₀ = (20/2)[2(5) + (20 - 1) * -1/2]
= 10[10 + 19 * -1/2]
= 10[10 - 19/2]
= 10[1/2]
= 5.
Therefore, Sum of first 20 terms of an AP is 115 (or) 5.
Hope it helps!
Answer:
Step-by-step explanation:
Here is your solution
Given:-
The sum of 3rd term and 7th term
=>a+2d+a+6d = 6
=>2a+8d = 6
After common taking
a + 4d = 3
a=3-4d ..................….......(1)
Their product is
(a+2d)×(a+6d) = 8 .........…(2)
Putting value of a = 3–4d in equation (2) to get
=>(a+2d)×(a+6d) = 8
=>(3–4d+2d)×(3–4d+6d) =8
=>(3–2d)×(3+2d) = 8
=>9–4d^2 = 8
=>4d^2 = 9-8
=>4d^2 = 1
=>d^2 = 1/4
=>d = 1/2.
From (1) a = 3–4d
a=> 3–4×1/2
a=>3-2
a=> 1
The sum of the first 16 terms
Sum of 16 terms = (20/2)[2a+(20-1)d]
= (20/2)[2×1 + (20–1)×1/2]
= 10[2+19/2]
= 10×23/2
= 230/2
=115
Hence,
The sum of first 20 term is 115
Hope it helps you