The sum of 7terms and 3rd term of an AP is 6 and their product is 8.find the sum of first 20th terms
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The sum of 7th term and 3rd term of an AP
we know that nth Term in AP = a + (n-1)d
7th Term = a +(7-1)d = a + 6d...(1)
3rd Term= a + (3 - 1)d = a + 2d...(2)
Given ''sum of (1) & (2) is 6
(a +6d) + (a + 2d) = 6
(2a + 8d) = 6
2(a + 4d) = 2(3)
a + 4d = 3....(3)
a = 3 - 4d....(4)
Given"product of (1) & 2 is 8"
(a + 6d) (a +2d) = 8
substitute (4)
(3 -4d + 6d) ( 3 -4d + 2d) = 8
(3 +2d) ( 3 -2d) = 8
its in the form of (a+b)(a-b) = (a^2 - b^2)
(3)^2 - (2d)^2 = 8
9 - 4d^2 = 8
9-8 = 4d^2
1/4 = d^2
√(1/4) = d
1/2 = d... (5) substitute it in 4
a = 3 -4d
a = 3 - 4(1/2)
a = 3 - 2
a = 1... (6)
sum of n terms= n/2[ 2a + (n-1)d]
sum of 20 terms
= 20/2 [2(1) + (20 - 1)1/2]
= 10[ 2 + 19/2]
= 10[2 + 9.5]
= 10[11.5]
= 115
we know that nth Term in AP = a + (n-1)d
7th Term = a +(7-1)d = a + 6d...(1)
3rd Term= a + (3 - 1)d = a + 2d...(2)
Given ''sum of (1) & (2) is 6
(a +6d) + (a + 2d) = 6
(2a + 8d) = 6
2(a + 4d) = 2(3)
a + 4d = 3....(3)
a = 3 - 4d....(4)
Given"product of (1) & 2 is 8"
(a + 6d) (a +2d) = 8
substitute (4)
(3 -4d + 6d) ( 3 -4d + 2d) = 8
(3 +2d) ( 3 -2d) = 8
its in the form of (a+b)(a-b) = (a^2 - b^2)
(3)^2 - (2d)^2 = 8
9 - 4d^2 = 8
9-8 = 4d^2
1/4 = d^2
√(1/4) = d
1/2 = d... (5) substitute it in 4
a = 3 -4d
a = 3 - 4(1/2)
a = 3 - 2
a = 1... (6)
sum of n terms= n/2[ 2a + (n-1)d]
sum of 20 terms
= 20/2 [2(1) + (20 - 1)1/2]
= 10[ 2 + 19/2]
= 10[2 + 9.5]
= 10[11.5]
= 115
Answered by
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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 20 terms
Sum of 16 terms = (n/2)[2a+(n-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
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 20 terms
Sum of 16 terms = (n/2)[2a+(n-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
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