Question 1096859
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Plug in n = 1 to find the first term


{{{a[n] = n(n+9)}}}


{{{a[1] = 1(1+9)}}} n is replaced with 1


{{{a[1] = 1(10)}}}


{{{a[1] = 10}}}


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Plug in n = 2 to find the second term


{{{a[n] = n(n+9)}}}


{{{a[2] = 2(2+9)}}} Substitute 2 for n


{{{a[2] = 2(11)}}}


{{{a[2] = 22}}}


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Plug in n = 3 to find the third term


{{{a[n] = n(n+9)}}}


{{{a[3] = 3(3+9)}}}


{{{a[3] = 3(12)}}}


{{{a[3] = 36}}}


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Subtract the second and first terms:


{{{d[1] = a[2] - a[1] = 22 - 10 = 12}}}


Now subtract the second and third terms


{{{d[2] = a[3] - a[2] = 36-22 = 14}}}


Since {{{d[1] <> d[2]}}} this means we do not have an arithmetic sequence. 
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