The 9th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times is 15th term .

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Solution

Let a be the first term
Let d - common difference

$T_{9}$ = 2 $T_{10}$
a + 8d = 2(a + 9d )
a+8d=2a+18d
a = -10d

Here remember that this is relation is followed throughout the AP.
So, $T_{72}$ = 4 $T_{15}$
a + 71d = 4a + 60d
3a = 15d
a = 5d
so question asked is wrong

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