the sum of 4th and 8th term is 24the sum of the 6th and 10th term is 44 find the first three terms of the ap?
Answers
A8=a+7d
And
A6=a+5d
A10=a+9d
Then
a+3d+a+7d=24
a+5d+a+9d=44
By solving we get
(1)-- 2a+10d=24
(2)--2a+14d=44
By elimination we want to do that
We get d=5
Sub d value in (1) then
We get a=-13 then do
A1=a=-13
A2=a+d=-13+(5)=-8
A3=a+2d=-13+2(5)=-13+10=-3
Given:
On adding the 4th and the 8th term, the sum is 24
The sum of the 6th and 10th terms of A.P is 44.
To find:
The value of the first three terms of an A.P.
Solution:
Let us take the first three terms of an A.P as
a, (a+d), (a+2d)
(a) = first term
(a+d) = second term
(a+2d) = third term
d = common difference
The nth term of an A.P is calculated by the formula,
= a + (n-1) d
The 4th term is,
a₄ = a + ( 4-1) d ( n= 4)
a + 3d = a₄ ( equation 1)
The 8th term is,
a₈ = a+ (8-1) d ( n= 8)
a+7d = a₈ ( equation 2)
Since the 4th and 8th terms give 24 as the sum,
a₄ +a₈ = 24
Adding equation 1 ad equation 2,
a +3d + a +7d = 24
2a + 10d = 24
a + 5d = 12 ( equation 3)
The 6th term is,
a₆ = a + (6-1) d ( n = 6)
a +5d = a₆ ( equation 4 )
The 10th term is,
a₁₀ = a + (10-1) d ( n = 10)
a +9d = a₁₀ ( equation 5 )
Since 6th and 10th terms gives 44 as the sum,
a₆ + a₁₀ = 44
Adding equation 4 and equation 5,
a+5d + a+9d = 44
2a + 14d = 44
a + 7d = 22 ( equation 6)
Subtracting equations 3 from equation 6 ,we get,
2d = 10
d = 5
From equation 3,
A + (5× 5) = 12
A + 25 = 12
A = - 13
First term, a = -13
Second term, a+d = -8
Third term, a+2d = -3
The first three terms of the A.P are -13, -8, and -3.