# maths

the third term of an AP is 5 while the 7th term is 13 find the 30th term of the AP

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1. 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

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2. 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

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3. Don't know..sorry

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