Question

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

Answer 1

Hello,

Solution:

ATQ

Tn
$= a + (n - 1)d$
T4 = a+3d = 0

d= -a/3

25 term

T25 = a+24d

T25 = a+24(-a/3)

T25 = -7a .........eq1

T11 = a+10 d

= a+10(-a/3)

T11= -7a/3………eq2

on comparing eq1 and 2 ,we can say that 25th term of the A.P. is three times its 11th term

T11 (3) = T25

Hope it helps you