The 4th term of an arithmetic progression AP is 37 and the 6th term is 12 more than the 4th term find. (i)the first term of AP. (ii) the seventh term of AP.
Y=sinx
To find the first term and the seventh term of an arithmetic progression (AP), we can use the formula for the nth term of an AP:
nth term (Tn) = a + (n-1)d,
where a is the first term and d is the common difference.
(i) Finding the first term (a):
Given that the 4th term (T4) is 37, we can substitute n = 4 into the formula:
37 = a + (4-1)d
37 = a + 3d
(ii) Finding the sixth term (T6):
Given that the 6th term (T6) is 12 more than the 4th term (T4), we can substitute n = 6 and n = 4 into the formula:
T4 = a + (4-1)d
T6 = a + (6-1)d
Since T6 is 12 more than T4, we can rewrite the equation as:
T6 = T4 + 12
Now we can substitute the expressions for T4 and T6 using the formulas:
a + 3d = 37
a + 5d = 37 + 12
Simplifying the equations, we have:
a + 3d = 37 ----(1)
a + 5d = 49 ----(2)
Now we have a system of two equations that we can solve simultaneously to find the values of a and d.
First, we can subtract equation (1) from equation (2) to eliminate a:
(a + 5d) - (a + 3d) = 49 - 37
2d = 12
d = 6
Substituting the value of d back into equation (1):
a + 3(6) = 37
a + 18 = 37
a = 37 - 18
a = 19
Therefore, the first term (a) of the AP is 19.
Now we can find the seventh term (T7) of the AP by substituting n = 7 into the formula:
T7 = a + (7-1)d
T7 = 19 + (7-1)6
T7 = 19 + 6(6)
T7 = 19 + 36
T7 = 55
Therefore, the seventh term of the AP is 55.
Use your definitions:
The 4th term of an arithmetic progression AP is 37
---> a + 3d = 37 **
the 6th term is 12 more than the 4th term
---> a+5d - (a+3d) = 12
but a+3d = 37
a+5d - 37 = 12
a+5d = 49 ***
solve using ** and ***