Question

How many terms of the series 2 + 4 + 6 + ....must be taken in order so that the sum may be 420?

Answer 1

$\huge{\underline{\red{\fcolorbox{red}{pink}{Answer}}}}$

$\mathbf{\underline{given}}$

⤇series = 2,4,6....Tn

⤇ a = 2

⤇ d = 2

⤇ Sn = 420

$\mathbf{\underline{solution}}$

$\mathbf{\underline{Formula = n/2[ 2a+(n-1)d ]}}$

⇝ 420 = n/2 [ 2(2) + (n-1)2 ]

⇝ 840 = n [ 4+2n-2 ]

⇝ 840 = n[2+2n]

⇝ 840 = 2n +2n²

⇝ 0 = n² + n -420

⇝ 0 = (n+21) (n-20)

$\bf{{n \: = \: 20}}$

⇝ sum of 20 terms of this series is 420

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