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

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
11th term is a+10d

that will make a difference.