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
the third term of an AP is 5 while the 7th term is 13 find the 30th term of the AP
3 answers
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
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
Don't know..sorry
2 answers