Asked by Anonymous
the first three terms of an ap are x, (3x +1) and (7x -4). Find the value x and 10th term
Answers
Answered by
Bosnian
In A.P.
a1 = initial term of arithmetic progression
d = common difference
nth term of A.P.
an = a1 + ( n - 1 ) d
In this case a1 = x , a2 = 3 x + 1 , a3 = 7 x - 4 so:
a1 = x
a2 = a1 + ( 2 - 1 ) d
a2 = a1 + d
3 x + 1 = x + d
a3 = a1 + ( 3 - 1 ) d
a3 = a1 + 2 d
7 x - 4 = x + 2 d
Now you must solve system:
3 x + 1 = x + d
7 x - 4 = x + 2 d
Ttry it.
The solutions are:
x = 3 , d = 7
a1 = x = 3
an = a1 + ( n - 1 ) d
a10 = a1 + ( 10 - 1 ) d
a10 = a1 + 9 d
a10 = 3 + 9 ∙ 7 = 3 + 63 = 66
Your A.P.
3 , 10 , 17 , 24 , 31 , 38 , 45 , 52 , 59 , 66 ...
Proof:
a1 = x = 3
a2 = 3 x + 1
10 = 3 ∙ 3 + 1 = 9 + 1 = 10
a3 = 7 x - 4
17 = 7 ∙ 3 - 4 = 21 - 4 = 17
a1 = initial term of arithmetic progression
d = common difference
nth term of A.P.
an = a1 + ( n - 1 ) d
In this case a1 = x , a2 = 3 x + 1 , a3 = 7 x - 4 so:
a1 = x
a2 = a1 + ( 2 - 1 ) d
a2 = a1 + d
3 x + 1 = x + d
a3 = a1 + ( 3 - 1 ) d
a3 = a1 + 2 d
7 x - 4 = x + 2 d
Now you must solve system:
3 x + 1 = x + d
7 x - 4 = x + 2 d
Ttry it.
The solutions are:
x = 3 , d = 7
a1 = x = 3
an = a1 + ( n - 1 ) d
a10 = a1 + ( 10 - 1 ) d
a10 = a1 + 9 d
a10 = 3 + 9 ∙ 7 = 3 + 63 = 66
Your A.P.
3 , 10 , 17 , 24 , 31 , 38 , 45 , 52 , 59 , 66 ...
Proof:
a1 = x = 3
a2 = 3 x + 1
10 = 3 ∙ 3 + 1 = 9 + 1 = 10
a3 = 7 x - 4
17 = 7 ∙ 3 - 4 = 21 - 4 = 17
Answered by
Ayodeji
please solve it clear
Answered by
micheal
[(7 x - 4 = x + 2 d) - (3 x + 1 = x + d)] => 4 x - 5 = d
substitute the value of d(4 x - 5) into any of the equations (7 x - 4 = x + 2 d) or (3 x + 1 = x + d)
substitute the value of d(4 x - 5) into any of the equations (7 x - 4 = x + 2 d) or (3 x + 1 = x + d)
Answered by
Esther
Nice work
Answered by
haneefarh
wow that's so brilliant
Answered by
ojaga
I dont understand how did got 63
Answered by
Nerdy...
X,(3x+1), and (7x-4)
Soln
Arithmetic mean: b=a+c ÷2
b=3x+1, a= x, c=7x-4
3x+1= x + 7x-4/2
Cross multiply
2(3x+1) = ×+7x-4
6x+2 = 8x-4
Collect like terms
6x-8x = -4-2
-2x =-2
Divide through with -2
Therefore x= 1
Soln
Arithmetic mean: b=a+c ÷2
b=3x+1, a= x, c=7x-4
3x+1= x + 7x-4/2
Cross multiply
2(3x+1) = ×+7x-4
6x+2 = 8x-4
Collect like terms
6x-8x = -4-2
-2x =-2
Divide through with -2
Therefore x= 1
Answered by
Nerdy...
The others are using the wrong formula to truly find the first three terms of an Ap the arithmetic mean's formula does the job
Answered by
Buzzy
Nice one I love it🤞
Answered by
Yomade
i am not sure I understand the answer
Answered by
Victory
I don't understand
Answered by
Chidera
I don't understand
Answered by
Chidera
What about the first seven terms of progression?
Answer
U
Answered by
Onuoha Emilia
Let T1=x
T2=3x+1
T3=7x-4
-T1=x
-T2=a1+(2-1)d
T2=a1+d
Substitute T2 in the eqn above
3x+1=x+d......(I)
-T3=a1+(3-1)d
T2=a1+2d
Substitute T3 in the eqn above
7x-4=x+2d.....(ii)
Equate both eqn to zero
- 3x+1-x-d=0
=2x+1-d=0.....(iii)
- 7x-4-x-2d=0
=6x-4-2d=0....(iv)
Eliminate d using eqn III & iv
2x+1-d=0...×2
6x-4-2d=0..×1
= 4x+2-2d=0
- 6x-4-2d=0
= -2x+6=0
-2x=-6
Divide both side by -2
X=3
Add xto eqn iv
6(3)-4-2d=0
18-4-2d=0
18-4=2d
14=2d
Divide both side by 2
7=d
- since T1=3
: Tn=3+ (10-1)7
= 3+9×7
= 3+63
=67.
Thanks 😊☺️😘
T2=3x+1
T3=7x-4
-T1=x
-T2=a1+(2-1)d
T2=a1+d
Substitute T2 in the eqn above
3x+1=x+d......(I)
-T3=a1+(3-1)d
T2=a1+2d
Substitute T3 in the eqn above
7x-4=x+2d.....(ii)
Equate both eqn to zero
- 3x+1-x-d=0
=2x+1-d=0.....(iii)
- 7x-4-x-2d=0
=6x-4-2d=0....(iv)
Eliminate d using eqn III & iv
2x+1-d=0...×2
6x-4-2d=0..×1
= 4x+2-2d=0
- 6x-4-2d=0
= -2x+6=0
-2x=-6
Divide both side by -2
X=3
Add xto eqn iv
6(3)-4-2d=0
18-4-2d=0
18-4=2d
14=2d
Divide both side by 2
7=d
- since T1=3
: Tn=3+ (10-1)7
= 3+9×7
= 3+63
=67.
Thanks 😊☺️😘
Answered by
B
don't understand either
Answer
Grace
Answered by
Gold
The first three terms of an A.P are 1,(3×+1) and (7×-4). find the 10th term
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