the third term of an AP is 5 while the 7th term is 13 find the 30th term of the AP
T3 and T7 are 4 terms apart, so if the common difference is d, then 4d=13-5, so d=2.
Now T30 is 23 terms past T7, so add 23*2 to T7 and you get T30=59
Don't know..sorry
To find the 30th term of an Arithmetic Progression (AP) when you are given the third term and the seventh term, we can use a formula.
The formula to find the nth term in an AP is:
An = A + (n - 1)d
Here, An represents the nth term of the AP, A is the first term, n is the position of the term you want to find, and d is the common difference between terms.
In this case, we have the third term as 5 and the seventh term as 13.
Let's use these values to find the common difference (d) first:
A3 = A + (3 - 1)d
5 = A + 2d
A7 = A + (7 - 1)d
13 = A + 6d
We have two equations with two unknowns (A and d). We can solve them simultaneously.
Subtracting the first equation from the second equation:
13 - 5 = (A + 6d) - (A + 2d)
8 = 4d
Dividing both sides by 4:
d = 8/4
d = 2
Now that we have the common difference, we can find the first term (A) by substituting the value of d into either of the original equations:
5 = A + 2(2)
5 = A + 4
A = 5 - 4
A = 1
So, the first term (A) is 1, and the common difference (d) is 2.
Now, we can use the formula to find the 30th term:
A30 = 1 + (30 - 1)2
A30 = 1 + 29(2)
A30 = 1 + 58
A30 = 59
Therefore, the 30th term of the AP is 59.
Third term a+2d is 5
7th. Term a+6d is 13
By solving using elimination method,
-4d= 8
d=-2
Substitute d value to third term
a-4 = 5
a= 9
Therefore 30th term is a + 29d = 9+29(-2) = 9-58 = -49