Q. 125.0( 6 Votes )

# In a certain A.P. the 24^{th} term is twice the 10^{th} term. Prove that the 72^{nd} term is twice the 34^{th} term.

Answer :

Given: 24^{th} term is twice the 10^{th} term

⇒ a_{24} = 2a_{10}

To prove: a_{72} = 2a_{34}

We know, a_{n} = a + (n – 1)d where a is first term or a_{1} and d is common difference and n is any natural number

When n = 10:

∴ a_{10} = a + (10 – 1)d

⇒ a_{10} = a + 9d

When n = 24:

∴ a_{24} = a + (24 – 1)d

⇒ a_{24} = a + 23d

When n = 34:

∴ a_{34} = a + (34 – 1)d

⇒ a_{34} = a + 33d ………(i)

When n = 72:

∴ a_{72} = a + (72 – 1)d

⇒ a_{72} = a + 71d

According to question:

a_{24} = 2a_{10}

⇒ a + 23d = 2(a + 9d)

⇒ a + 23d = 2a + 18d

⇒ a – 2a + 23d – 18d = 0

⇒ -a + 5d = 0

⇒ a = 5d

Now, a_{72} = a + 71d

⇒ a_{72} = 5d + 71d

⇒ a_{72} = 76d

⇒ a_{72} = 10d + 66d

⇒ a_{72} = 2(5d + 33d)

{∵ a = 5d}

⇒ a_{72} = 2(a + 33d)

⇒ a_{72} = 2a_{34} (From (i))

Hence Proved

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