the 9th term of an AP is three times the 5th term. find a relation between a and d and prove that the 8th term is 3 times the 4th term.
Tn = a + (n-1) d
T9 = a + 8 d
T5 = a + 4 d
so
a+8 d = 3a + 12 d
-4 d = 2 a
d = - a/2
T8 = a + 7 d = a - 7 a /2 = -5 a/2
T4 = a + 3 d = a - 3 a/2 = -1 a/2
I get 5 times. Check my arithmetic
"the 9th term of an AP is three times the 5th term" ---> a+8d = 3(a+4d)
a+8d = 3a + 12d
-2a = 4d
a = -2d
term(8) = a + 7d = -2d + 7d = 5d
term(4) = a + 3d = -2d + 3d = d
So the 8th term is 5 times the 4th term, not 3 times
e.g. let d = -3
then a = 6, and the sequence is
6, 3, 0, -3, -6, -9, -12, -15, -18, ...
9th term is -18, the 5th term is -6, so the 9th term is 3 times the 5th, check!
8th term is -15, 4th term is -5, so the 8th term is 5 times the 4th term
looks like you have a typo, or else the question is bogus