find the first three terms of a sequence whose nth term is given by tn=2n+1
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Answered by
14
Tn = 2n +1
=> a+(n-1)d = 2n - 2 + 3
=> a+(n-1)d = 3 + 2(n-1)
On Comparing both sides, we get
a = 3
d = 2
T1 = 3
T2 = a +d = 3 + 2 = 5
T3 = a +2d = 3 + 2*2 = 7
First three terms are 3,5 and 7
=> a+(n-1)d = 2n - 2 + 3
=> a+(n-1)d = 3 + 2(n-1)
On Comparing both sides, we get
a = 3
d = 2
T1 = 3
T2 = a +d = 3 + 2 = 5
T3 = a +2d = 3 + 2*2 = 7
First three terms are 3,5 and 7
Answered by
10
Since nth term is given by tn=2n+1
first term = t1 = 2×1+1 = 3
second term = t 2 = 2×2 +1 = 5
third term = t3 = 2×3 +1 =7
first term = t1 = 2×1+1 = 3
second term = t 2 = 2×2 +1 = 5
third term = t3 = 2×3 +1 =7
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