The sum of $20$ terms of an A.P. whose nth term $4 n - 1$ is :

820

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79

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80 n - 1

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80 n + 1

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Solution

The correct option is A $820$
Given, $a_{n} = 4 n - 1$ putting $n = 1, 2, 3, . . . . . .$
$a_{1} = 3$
$a_{2} = 7$
$a_{3} = 11$
$a_{4} = 15$
$∴$ Series is $3, 7, 11, 15, . . . . . .$
$S_{20} = \frac{20}{2} {2 (3) + (20 - 1) 4}$
$= 10 {6 + 76}$
$= 10 \times 82$
$= 820$

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