The fourth term of an A.P. is ten times the first. Prove that the sixth term is four times as greater as second term.

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Solution

Let first term = a & common diff = d
Fourth term = a+3d
$∴ a + 3 d = 10 a$ or 3d=9a
or d=3a
2nd term = a+b = a+3a = 4a
6th term = a+5d = a+15a = 16a
$∴$ 6th term = 4 $\times$ 4 a
Hence, 6th term = 4 $\times$ 2nd term

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The fourth term of an A.P. is ten times the first. Prove that the sixth term is four times as greater as second term.

Let first term = a & common diff = d Fourth term = a+3d ∴a+3d=10a or 3d=9a or d=3a 2nd term = a+b = a+3a = 4a 6th term = a+5d = a+15a = 16a ∴ 6th term = 4 × 4 a Hence, 6th term = 4 × 2nd term

Let first term = a & common diff = d
Fourth term = a+3d
$∴ a + 3 d = 10 a$ or 3d=9a
or d=3a
2nd term = a+b = a+3a = 4a
6th term = a+5d = a+15a = 16a
$∴$ 6th term = 4 $\times$ 4 a
Hence, 6th term = 4 $\times$ 2nd term