miscellaneous question of class 11 chapter 9 question 22
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The First AP is given as follows: -
2, 4, 6…
where, first term(a) = 2
common difference(d) = 4 - 2 = 2
∴
nth 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
∴
nth term = a + (n - 1)d
= 4 + (n - 1)2
= 4 + 2n - 2
= 2n + 2
Now,
an = [nth term of 2, 4, 6…] × [nth term of 4, 6, 8…]
= (2n) × (2n + 2)
= 4n2 + 4n
Thus, the nth term of series 2 × 4 + 4 × 6 + 6 × 8 + ... is
an = 4n2 + 4n
∴ a20 = 4 × (20)2 + 4 × 20 = 1600 + 80 = 1680
Hence, 20th term of series is 1680.
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