The fourth term of an ap is zero. Prove that 25th term of the ap is three times its 11th term
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t(4) =0
a+3d =0
t(25)
= a + 24d
= - 3d + 24d
= 21d
= 3 × 7d
= 3 × (a+7d-a)
= 3 × (a+7d+3d)
= 3 × (a+10d)
= 3× t(11)
Hence proved
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a+3d =0
t(25)
= a + 24d
= - 3d + 24d
= 21d
= 3 × 7d
= 3 × (a+7d-a)
= 3 × (a+7d+3d)
= 3 × (a+10d)
= 3× t(11)
Hence proved
Mark this answer as brainliest answer
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