Question

If the sum of first n even natural numbers is 420 . Find the value of n

Answer 1

HEYA !

Given that,

Sum of first n even natural numbers
= 420

(2 + 4 + 6 + . . . + 2n) = 420

2 (1 + 2 + 3 + . . . + n) = 420

2 [ n (n + 1) / 2 ] = 420

n (n + 1) = 420

n (n + 1) = 21 × 22

n (n + 1) = 21 × (21 + 1)

By comparing , We get

$\boxed{ \: \text{n = 21} \: } \: \: \: \: \: \: \mathbf{Ans.}$
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Formula Used :

(1 + 2 + 3 + . . . + n) = [ n(n + 1) / 2 ]

Answer 2

We have,

2 + 4 + 6 + 8 + ........... to n terms = 420

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We know that,

$\star\;{\boxed{\sf{\pink{S_n = \dfrac{n}{2}\bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup}}}}$

Putting values,

$\sf Here \begin{cases} & \sf{First\;term,\;a = \bf{2}} \\ & \sf{Common\; difference,\;d = \bf{2}} \\ & \sf{S_n = \bf{420}} \end{cases}$

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$:\implies\sf 420 = \dfrac{n}{2}\bigg\lgroup\sf 2 \times 2 + (n - 1) \times 2 \bigg\rgroup$

$:\implies\sf 420 = \dfrac{n}{2}\bigg\lgroup\sf 4 + 2n - 2 \bigg\rgroup$

$:\implies\sf 420 = \dfrac{n}{2}\bigg\lgroup\sf 2n + 2\bigg\rgroup$

$:\implies\sf 420 \times 2 = n\bigg\lgroup\sf 2n + 2\bigg\rgroup$

$:\implies\sf 840 = 2n^2 + 2n$

$:\implies\sf 840 = 2(n^2 + n)$

$:\implies\sf \cancel{ \dfrac{840}{2}} = n^2 + n$

$:\implies\sf 420 = n^2 + n$

$:\implies\sf n^2 + n - 420 = 0$

$:\implies\sf n^2 + 21n - 20n - 420 = 0$

$:\implies\sf n(n + 21) - 20(n - 21) = 0$

$:\implies\sf (n + 21)(n - 20) = 0$

$:\implies\sf n = 20, -21$

$:\implies{\boxed{\frak{\purple{n = 20}}}}\;\bigstar\qquad\qquad\bigg\lgroup\sf \because\;n\;is\;a\;natural\;no. \bigg\rgroup$

$\therefore\;{\underline{\sf{Sum\;of\;20\; natural\;numbers\;is\;420.}}}$