Q. 274.3( 24 Votes )

# The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.

Answer :

Let *a* be the first term and *d* be the common difference.

Given: a_{4} = 0

To prove: a_{25} = 3 × a_{11}

Now, Consider a_{4} = 0

⇒ a + (4 - 1)d = 0

⇒ a + 3d = 0

⇒ a = - 3d ………. (1)

Consider a_{25} = a + (25 - 1)d

⇒ a_{25} = - 3d + 24d (from equation (1))

⇒ a_{25} = 21d ………. (2)

Now, consider a_{11} = a + (11 - 1)d

⇒ a_{11} = - 3d + 10d (from equation (1))

⇒ a_{11} = 7d ……….(3)

From equation (2) and (3), we get,

a_{25} = 3 × a_{11}

Hence, proved.

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Find the indicated terms in each of the following arithmetic progression:

a = 3, d = 2; ; t_{n}, t_{10}

Find the indicated terms in each of the following arithmetic progression:

a = 21, d = — 5; t_{n}, t_{25}

Find the indicated terms in each of the following arithmetic progression: