The fourth term of an AP is 0. Prove that its 25th term is triple its 11th term.
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Given that, fourth term, a₄ = 0
As a₄ = a+(4-1)d
hence, a₄ = a+3d = 0
hence, a = -3d -------------1
To prove, 25th term = 3 times 11th term
hence to prove, a+(25-1)d = 3[a+(11-1)d]
solving this further, a+24d = 3[a+10d]
solving this further, a+24d = 3a+30d
LHS = a+24d
substituting from 1, a = -3d in LHS, we get
LHS = -3d+24d = 21d
RHS = 3a+30d
substituting from 1, a = -3d in RHS, we get
RHS = 3(-3d)+30d = -9d+30d = 21d
Clearly, LHS = RHS
Hence proved...
As a₄ = a+(4-1)d
hence, a₄ = a+3d = 0
hence, a = -3d -------------1
To prove, 25th term = 3 times 11th term
hence to prove, a+(25-1)d = 3[a+(11-1)d]
solving this further, a+24d = 3[a+10d]
solving this further, a+24d = 3a+30d
LHS = a+24d
substituting from 1, a = -3d in LHS, we get
LHS = -3d+24d = 21d
RHS = 3a+30d
substituting from 1, a = -3d in RHS, we get
RHS = 3(-3d)+30d = -9d+30d = 21d
Clearly, LHS = RHS
Hence proved...
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