The given 5, 3, 1,…………..is an arithmetic progression here
a = 5, d = a2 – a2 = 3 – 5 = -2
Let -25 is some of ‘n’ th term
i.e. an – 25
So an = a + (n-1)d
-25 = 5 + (n – 1)(- 2)
– 25 – 5 = (n – 1) (-2)
\(\frac{-30}{-2}\) = n – 1
⇒ n – 1 = 15 and n = 15+ 1 = 16
So – 25 exist at 16th term in above series.