The sum of first n terms of an AP is given Sn=2nsquare+3n.find 16 term of an AP
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of the AP -
when n=1
ist term = 2(1)^2 + 3(1) = 2 + 3 = 5
when n=2
sum of 1st and 2nd terms = 2(2)^2 + 3(2) = 8+6=14
2nd term = 14 - 5 = 9
common difference = 9-5 = 4
16th term = a+(n-1)d
= 5 + (16-1)(4)
= 5 + 60
= 65
when n=1
ist term = 2(1)^2 + 3(1) = 2 + 3 = 5
when n=2
sum of 1st and 2nd terms = 2(2)^2 + 3(2) = 8+6=14
2nd term = 14 - 5 = 9
common difference = 9-5 = 4
16th term = a+(n-1)d
= 5 + (16-1)(4)
= 5 + 60
= 65
sagar2k:
dont u think it's wrong
means??
am I wrong??
yea I was wrong im sorry I wasn't careful
it's OK yrr
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OK bro
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