Question

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The sum of the first n consecutive even numbers can be found using s=n^2+n,where n≥2.
What is the value of n when the sum is 56?

A 6
B 39
C 26
D 12

show your work

Answer 1

Answer:

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Answer 2

We have been given that the sum of 'n' terms of an AP can be found using the expression:-

$S_{n}=n^{2}+n$

We have been asked to find the value of 'n' when this expression is equal to 56.

According to question:-

$n^{2} + n = 56$

$n^{2} + n - 56 = 0$

Middle Term Splitting:-

$n^{2} + 8n - 7n - 56 = 0$

$n(n+8) - 7(n+8) = 0$

$(n+8)(n-7) = 0$

We get two values of n as -8 and 7.

-8 would be rejected as the number of terms can't be negative.

So, the value of 'n' is 7 when the sum is 56 !