Question

the 4th term of an A.P is zero.Prove that its 25th term is triple its 11th term

Answer 1

let in an first term =a , common difference = d
nth term = tn = a+(n-1)d

given t4=0
i) a+3d=0 ⇒a = -3d ---(1)

ii)t11 =a+10d
t11=-3d+10d [ from (1)]
=7d--(2)
iii) t25 = a+24d

= -3d+24d [ from (1)]
= 21d
=3*7d
t25=3*t11
therefore
25th term of an AP = 3 times of its 11th term