Find the value of n,if the sum to n terms of the series √3+√75+√243+...is 435√3.
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sooooooooooory I can't help you
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Sn=435√3
a=√3
d=5√3-√3
=4√3
n=?
Sn=n/2[2a+(n-1)d]
435√3=n/2[2(√3)+(n-1)4√3]
435√3=n/2[2√3+4√3n-4√3]
870√3=n[4√3n-2√3]
4√3n^2-2√3n-870√3=0
FACTORISE FURTHUR TO GET THE VALUES OF N.
HOPE THIS HELPS U.
a=√3
d=5√3-√3
=4√3
n=?
Sn=n/2[2a+(n-1)d]
435√3=n/2[2(√3)+(n-1)4√3]
435√3=n/2[2√3+4√3n-4√3]
870√3=n[4√3n-2√3]
4√3n^2-2√3n-870√3=0
FACTORISE FURTHUR TO GET THE VALUES OF N.
HOPE THIS HELPS U.
Anonymous:
I want after this step only.
Plzszzs
But I've written in different way
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