how many terms of the series 18+15+12+.......,when added together will give 45?
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let no. of required term is n.
then...
Sn = n/2[2a+(n-1)d]
45 = n/2[2×18+(n-1)(-3)]
45=n/2[39-3n]
90=39n - 3n^2
3n^2 -39n +90=0
after solving...
n = 10 & 3
then...
Sn = n/2[2a+(n-1)d]
45 = n/2[2×18+(n-1)(-3)]
45=n/2[39-3n]
90=39n - 3n^2
3n^2 -39n +90=0
after solving...
n = 10 & 3
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