The Sum Of first m terms of an AP is 4m 2 -m if its nth term is 107
find the value of n & the 21st term of this AP
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Sn = 4m²-m
a1 = S1 = 4(1)²-(1) = 3
a2 = S2-S1 = 4(2)²-2-3 = 16-5 = 11
d = a2-a1 = 11-3 = 8
nth term is 107 (given)
Tn = 107
a1+(n-1)d = 107
3+(n-1)8 = 107
n-1 = 104/8 = 13
=> n = 14
Now,
20th term
T20 = a1+19d
= 3+19×8 = 3 + 152 = 155
a1 = S1 = 4(1)²-(1) = 3
a2 = S2-S1 = 4(2)²-2-3 = 16-5 = 11
d = a2-a1 = 11-3 = 8
nth term is 107 (given)
Tn = 107
a1+(n-1)d = 107
3+(n-1)8 = 107
n-1 = 104/8 = 13
=> n = 14
Now,
20th term
T20 = a1+19d
= 3+19×8 = 3 + 152 = 155
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