Question

the 4th term of an AP is zero. prove that the 25th term of an AP is three times its 11th term.

Answer 1

.here,
t4 = 0
let the first term of an ap be a and the common difference be x
therefore,
t4 = 0 =a + 3x
a =-3x....1
t11=a +10x
substituting the value of a from eqn1 we get,
t11 = -3x + 10x =7 x.....2
t25 = a + 24x
using eqn1 we get,
t25 = -3x + 24x = 21x=3(7x)....3
joining eqn2 and eqn2 we get,
t25 = 3(t11)
(proved)

Answer 2

kindly refers to the attached file for answer.



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