Here , a=50
d=50-45= -5
Let the nth term of the given AP be the first negative term .
Then ,\(a_n<0\)
\(=( {a+(n-1)d} )<0\)
\(=( {50+(n-1)(-5)} )<0\)
\(=( 50-5n+5)<0\)
\(=55-5n<0\)
\(=55<5n\)
\(=5n>55\)
\(=n>11\)
\(\therefore n=12\)
Hence , the 12th term is the first negative term of the given AP
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