which term of the ap 3,15,27,39....will be 120 moee than its 21 term
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Common difference of series = (n+1) – n
We get 15 – 3, 27 – 15……
Common difference = 12
First term of series = 3.
Therefore, the 21st term of the series is
T21=a+d(n-1)
T21=3+12(21-1)
t21=243
required term after adding 120=120+243=363
LET tn=363
first value of the series is a =3
common difference =12
n th term term is 363.
VALUE OF N.
tn =3+12(n-1)
363=3+12(n-1)
n=31
.
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