Find the values of P and Q in an Ap -12,P,Q,18
In an arithmetic progression;
a1
a2 = a1 + d
a3 = a1 + 2d
a4 = a1 + 3d
In this case:
a1 = -12
a2 = P
a3 = Q
a4 = 18
a4 - a1 = a1 + 3d - a1
a4 - a1 = 3 d
18 - ( - 12 ) = 3 d
18 + 12 = 3 d
30 = 3 d
3 d = 30
Divide both sides by 3
d = 10
a2 = P = a1 + d = - 12 + 10 = - 2
a3 = Q = a1 + 2 d = - 12 + 2 ∙ ( 10 ) = - 12 + 20 = 8
there are three jumps from-12 to 18, a distance of 30.Each jump 10.
-12,-2, 8,18, 28, ....
T1= a (-12)
T2= a+d (p)
T3= a+2d (q)
T4= a+3d (18)
T4 - T1 = a +3d - a
T4
Well, well, well. Looks like we have ourselves an arithmetic progression, huh? Let's solve this puzzle.
In an arithmetic progression, the common difference (d) between consecutive terms remains the same. We need to find the values of P and Q, so let me put on my thinking clown nose.
We have -12, P, Q, and 18. The common difference (d) between consecutive terms would be the same for all pairs. Let's go step by step.
From -12 to P, we add the same difference. From P to Q, we again add the same difference. And finally, from Q to 18, we add the same difference.
Now, the difference between P and -12 is P - (-12) = P + 12. Similarly, the difference between Q and P is Q - P, and the difference between 18 and Q is 18 - Q.
So, P + 12 = Q - P = 18 - Q.
From these three equations, we can solve for P and Q by adding the first and the third equation together and then rearranging:
(P + 12) + (18 - Q) = 0
Simplifying,
P + 12 + 18 - Q = 0
P + 30 = Q
Now, we can substitute this expression for Q into one of the previous equations to solve for P. Let's use Q - P = P + 12:
(P + 30) - P = P + 12
30 = P + 12
18 = P
So, P is 18. And using P + 30 = Q, we can find Q:
18 + 30 = Q
Q = 48
Ta-dah! The values of P and Q in the arithmetic progression -12, P, Q, 18 are P = 18 and Q = 48.
I hope my clown-ish approach to math made you smile!
To find the values of P and Q in the arithmetic progression (AP) -12, P, Q, 18, we need to use the formula for the nth term of an arithmetic progression.
The nth term formula for an arithmetic progression is given by:
T_n = a + (n - 1)d
where T_n is the nth term, a is the first term, n is the position of the term in the sequence, and d is the common difference between consecutive terms.
In our case, the first term, a, is -12, and the fourth term, T_4, is 18. We are trying to find the values of P and Q, which correspond to the second and third terms respectively.
Using the formula, we have:
T_2 = -12 + (2 - 1)d
P = -12 + d
T_3 = -12 + (3 - 1)d
Q = -12 + 2d
T_4 = -12 + (4 - 1)d
18 = -12 + 3d
Now we have a system of equations with three unknowns (-12, P, and Q) and one equation. We can solve this system to find the values of P and Q.
Substituting the equations for P and Q into the equation for T_4:
18 = -12 + 3d
30 = 3d
d = 10
Now substituting the value of d back into the equations for P and Q:
P = -12 + d
P = -12 + 10
P = -2
Q = -12 + 2d
Q = -12 + 2(10)
Q = -12 + 20
Q = 8
Therefore, the values of P and Q in the arithmetic progression -12, P, Q, 18 are P = -2 and Q = 8.