the first three terms of an ap are x, (3x +1) and (7x -4). Find the value x and 10th term
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
Nice work
please solve it clear
I dont understand how did got 63
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.
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[(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)