Question

the 24th term of an ap is twice its 10 term show that 72 term is 4 times its 15term

Answer 1

Heya !!!


24th term = 2( 10th term )

A + 23D = 2 ( A + 9D)


A + 23D = 2A + 18D


2A - A = 23D - 18D


A = 5D



To prove :

72 term = 4 ( 15th term )


A + 71D = 4 ( A + 14D )


A + 71D = 4A + 56D

5D + 71D = 4 × 5D + 56D


76D = 20D + 56D


76D = 76D.......PROVED......



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Answer 2

Hello friend ☺☺ ― Solution here
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Let a be the first term and d be the common difference of the A.P . Than

T24 = 2 T10 (Given)

=> a + (n - 1) d

=> {a + (24 - 1) d} = 2 × (a + 9d)

=> a + 23d = 2a + 18d

=> 2a - a = 23 - 18d

=> a = 5d. ........ ( 1 )

Now ,

$\frac{a72}{a15} = \frac{a + 71d}{a + 14d} \\ \\ = > \frac{a72}{a15} = \frac{5d + 71d}{5d + 14d} \: \: \: \: \: \: \: \: \: \: \: (from \: (1)) \\ \\ = > \frac{a72}{a15} = \frac{76d}{19d} = 4 \\ \\ = > a72 = 4 \times a15$
Hence , the 72nd term of the A.P is 4 times it's 15th term.

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Hope it's helps you.
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