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

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Solution

We know that

$T_{n} = a + (n - 1) d$

Given, $T_{4} = a + (4 - 1) d = 0$

$a + 3 d = 0$

$a = - 3 d$

$T_{25} = a + (25 - 1) d$

= $a + 24 d = (- 3 d) + 24 d$

= $21 d$

And, $T_{11} = a + (11 - 1) d$

= $a + 10 d$

Then, $3 T_{11} = 3 (a + 10 d)$

= $3 a + 30 d$

= $3 (- 3 d) + 30 d (a = - 3 d)$

= $30 d - 9 d = 21 d = T_{25}$

$∴ 3 T_{11} = T_{25}$ Hence Proved.

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