Find the sum of the first 24 terms of the list of numbers whose n th term is given by an=3+2n
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an=3+2n
let take n as 1
a1=3+2(1)
a1=5
now, n as 2
so a2=3+2(2)
a2=7
d=a2-a1=7-5=2
sum:
Sn=n/2(2a+(n-1)d)
s24=24/2(2×5+(24-1)2)
=12(10+(23)2)
=12(10+46)
=12×56
=672
I hope this will help u
mark as brainleast
let take n as 1
a1=3+2(1)
a1=5
now, n as 2
so a2=3+2(2)
a2=7
d=a2-a1=7-5=2
sum:
Sn=n/2(2a+(n-1)d)
s24=24/2(2×5+(24-1)2)
=12(10+(23)2)
=12(10+46)
=12×56
=672
I hope this will help u
mark as brainleast
madhu637257:
tomorrow solluren.......
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