Math, asked by Swaza3571, 9 months ago

How many terms of the series 51+54+57+....are required to get a sum of 810

Answers

Answered by khairnarsanskriti
9

a1 = 51

a2 = 54

d = a2-a1

= 54-51 = 3

s = 810

s = n/2[ 2a+(n-1)d ]

810 = n/2 [ 2(51) + (n-1)3 ]

810= n/2 [ 102 + 3n - 3 ]

810 = n/2 [ 3n + 99 ]

810 x 2 = 3n^2 + 99n

1620 = 3n^2 + 99n

3n^2 + 99n -1620 = 0

n^2 + 33n - 540 = 0

solve further you will get it

Similar questions