Q. 224.2( 24 Votes )

# Find the 20^{th} term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms.

Answer :

The given series is in the form of multiplication of two different APs.

So, the nth term of given series is equal to the multiplication of their nth term.

The First AP is given as follows: -

2, 4, 6…

where, first term(a) = 2

common difference(d) = 4 - 2 = 2

∴

n^{th} term = a + (n - 1)d

= 2 + (n - 1)2

= 2 + 2n - 2

= 2n

The Second AP is given as follows: -

4, 6, 8…

where, first term(a) = 4

common difference(d) = 6 - 4 = 2

∴

n^{th} term = a + (n - 1)d

= 4 + (n - 1)2

= 4 + 2n - 2

= 2n + 2

Now,

a_{n} = [n^{th} term of 2, 4, 6…] × [n^{th} term of 4, 6, 8…]

= (2n) × (2n + 2)

= 4n^{2} + 4n

Thus, the n^{th} term of series 2 × 4 + 4 × 6 + 6 × 8 + ... is

a_{n} = 4n^{2} + 4n

∴ a_{20} = 4 × (20)^{2} + 4 × 20 = 1600 + 80 = 1680

Hence, 20^{th} term of series is 1680.

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