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2 years ago

Please help me with this #7

Mathematics

1 answer:

Alex2 years ago

5 0

$\text{Given that,}\\\\\text{First term,}~ a_1 = -8~~~~...(i)\\\\\text{12th term,}~ a_{12} = -52~~~...(ii)\\\\\text{Common difference is d.}\\\\(ii)\div(i):\\\\\dfrac{a_{12}}{a_1} = \dfrac{-52}{-8}\\\\\\\implies \dfrac{a_1 + (12-1)d}{a_1} = \dfrac{52}8\\\\\\\implies \dfrac{-8+11d}{-8} = \dfrac{52}8\\ \\\\\implies -8+11d = \dfrac{52(-8)}8\\\\\\\implies -8+11d = -52\\\\\\\implies 11d = -44\\\\\\ \implies d = \dfrac{-44}{11} = -4$

$\\\text{86th term,} ~a_{86} = a+ (86-1)d \\\\~~~~~~~~~~~~~~~~~~~~=-8+85d\\\\~~~~~~~~~~~~~~~~~~~~=-8+85(-4)\\\\~~~~~~~~~~~~~~~~~~~~=-348\\\\\text{Hence, the 86th term is}~ -348$

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