Given that T7=23,Tn=43 and T2n=91.for an A.P.find a,d and n...help please
T7=a+6d=23
Tn=a+[n-1]d=43
T2n=a+[2n-1]=91
subtract equ 2 from equ 3 give
nd=48---equ 4
from equ2 nd=43+d-a
:- 48=43+d-a
5=d-a ---equ6 recall from equ1 a=23-6d put the value of a into equ 6
d=4 from equ4 nd=48 since d=4, n=12
put the value of d into equ1 :- a=-1
(n=12,d=4,a=-1]
Thank you
What happened to the d in equation 3
I don't understand how you work it
You didn't include the d in equation 3
Why?
To find the common difference (d) and the number of terms (n) in an arithmetic progression (A.P.), we can use the given information about the terms T7, Tn, and T2n.
1. Determine the value of d:
Since T7 = 23, we know that T7 = a + 6d (where a is the first term).
Substitute the known values: 23 = a + 6d. Equation (1)
2. Determine the value of n:
Since Tn = 43, we know that Tn = a + (n - 1)d.
Substitute the known values: 43 = a + (n - 1)d. Equation (2)
3. Determine the value of T2n:
Since T2n = 91, we know that T2n = a + (2n - 1)d.
Substitute the known values: 91 = a + (2n - 1)d. Equation (3)
Now we have three equations (Equations 1, 2, and 3) with three unknowns (a, d, and n). We can solve these equations simultaneously to find the values of a, d, and n.
4. Solving the equations:
First, subtract Equation (2) from Equation (1) to eliminate a:
23 - 43 = (a + 6d) - (a + (n - 1)d)
-20 = 6d - (n - 1)d
-20 = 6d - nd + d
-20 = (6 - n)d + d
-20 = (7 - n)d. Equation (4)
Next, subtract Equation (2) from Equation (3) to eliminate a:
91 - 43 = (a + (2n - 1)d) - (a + (n - 1)d)
48 = 2nd - (n - 1)d
48 = 2nd - nd + d
48 = (2n - n)d + d
48 = (n)d + d
48 = (n + 1)d. Equation (5)
Now we have two equations (Equations 4 and 5) with two unknowns.
5. Solving the new equations:
Divide Equation (5) by Equation (4) to find the value of n:
(n + 1)d / ((7 - n)d) = 48 / -20
(n + 1) / (7 - n) = -48 / -20
(n + 1) / (7 - n) = 12 / 5
Cross-multiply and simplify the equation:
5(n + 1) = 12(7 - n)
5n + 5 = 84 - 12n
17n = 79
n = 79 / 17
n ā 4.65
Since the value of n is not a whole number, it implies that there is no solution for the given values of T7, Tn, and T2n in an arithmetic progression. Please double-check the given information or provide additional details to correct any errors or inconsistencies.
a+6d=23
a+[n-1]d=43
a+[2n-1]d=91
subtract #2 from #3 to get
nd = 48
Now the first two equations become
a+6d = 23
a+48-d = 43
subtract to get 7d = 28, so
d = 4, so a = -1, n=12