Find the sum of the first 15 terms of the series whose n th term is (4n + 1)?
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it is given that
a+(n-1)d=4n+1
a=4n+1-(n-1)d
a=4n+1-nd+d
now
s15=15by2(2a+14d)
s15=15 by2(2(4n+1-nd+d)+14d
=15by2(8n+2-2nd+2d+14d)
=15by2(8n+2-2nd+16d)
=15by2(120+2-30d+16d)
=15by2(122-14d)
=915-105d
a+(n-1)d=4n+1
a=4n+1-(n-1)d
a=4n+1-nd+d
now
s15=15by2(2a+14d)
s15=15 by2(2(4n+1-nd+d)+14d
=15by2(8n+2-2nd+2d+14d)
=15by2(8n+2-2nd+16d)
=15by2(120+2-30d+16d)
=15by2(122-14d)
=915-105d
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