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