Question 354971
since the difference between any two numbers is constant, that defines this sequence as arithmetic.
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The {{{nth}}} term is defined as {{{a[n]=a[1]+delta*(n-1)}}}
where {{{a[1]=1}}} (ie the first term and {{{delta}}}=difference between consecutive terms  {{{delta=3}}} in this sequence
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so...   {{{a[n]=a[1]+delta*(n-1)=1+3(n-1)=3n-2}}}
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sum of two consecutive terms:  {{{a[n]+a[n+1]=(3n-2)+(3(n+1)-2)=299}}}
therefore
3n-2+3n+3-2=299
6n-1=299
6n=300
n=50
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{{{a[n]=3n-2}}}=3*(50)-2=148
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{{{a[n+1]=3(n+1)-2=3(51)-2=151}}}
==
validate

148+151=299