thord te ?????
"The eleventh term of an A.P.with positive terms is four times its third term"
---> a+11d = 4(a+2d)
a+11d = 4a + 6d
5d = 3a
d = 3a/5
"the product of thord te and eleventh term is 256"
I will read that as third term
(a+2d)(a+11d) = 256
a^2 + 13ad + 11d^2 = 256
a^2 + 13a(3a/5) + 11(9a^2/25) = 256
a^2 + 39a^2/5 + 99a^2/25 = 256
times 25
25a^2 + 180a^2 + 99a^2 = 6400
304a^2 = 6400
a^2 = 6400/304 = 400/19
a = ± 20/√19
d = ±(3/5)(20/√19) = ±12/√19
AP is 20/√19 , 32/√19, 44/√19, ...
once you fix your typo, make any necessary changes to my solution
The eleventh term of an A.P.with positive terms is four times its third term and the product of thord te and eleventh term is 256.Find the A.P.
2 answers
11th term is a+10d
that will make a difference.
that will make a difference.