Arithmetic progressions 3

From the question,

$\dfrac{T_8 - T_4}{T_{15}} = \dfrac{4}{15}\\ (T_4 - T_1)^2 = 225$

Now, we let the first term be $a$ , and the common difference be $d$ .

$\dfrac{(a+7d) - (a+3d)}{a+14d} = \dfrac{4}{15}\\ 15(4d) = 4(a+14d)\\ 60d = 4a + 56d\\ 4a=4d\\ a=d$

From the second equation:

$((a+3d)-a)^2 = 225\\ 9d^2=225\\ d^2 = 25\\ d= \pm 5$

Now, if $d=-5,\;a=-5$ . Given that $d>0$ , we reject $d=-5$ .

Therefore, $d=5,\;a=5$

Now, we want to find the value of $n$ such that $T_n = 2015$

$5+(n-1)(5) = 2015\\ 5n=2015\\ n=\boxed{403}$

(a4-a1)^2 = 225 which means a4-a1 = 15 and so a4 = a1 + 15

Since a(n) = a1 + d* (n-1) then 15 = d * (3) yielding d = 5

(a8-a4)/a15 = 4/15. Cross multiplying gives 4 a15 = 15 a8 - 15*a4

a15 = a1 + 5 * 14 and a8 = a1 + 5 * 7

Substituting: 4 * (a1 + 70) = 15*(a1 + 35) - 15 * (a1 + 15)

Solving yields a1 = 5

2015 = a1 + d * (n - 1) so 2015 = 5 + 5 * (n-1)

2010 = 5 * (n-1)

402 = n -1

403 = n