### Find the greatest value of number of terms such that the sum of an increasing arithmetic sequence does not exceed a certain number

Question Sample Titled 'Find the greatest value of number of terms such that the sum of an increasing arithmetic sequence does not exceed a certain number'

For any positive integer ${n}$ , let ${A}{\left({n}\right)}={12}{n}+{8}$ and ${B}{\left({n}\right)}={10}^{{{12}{n}+{8}}}$ .

 (a) Express ${A}{\left({1}\right)}+{A}{\left({2}\right)}+{A}{\left({3}\right)}+\ldots+{A}{\left({n}\right)}$ in terms of ${n}$ . (2 marks) (b) Find the greatest value of ${n}$ such that ${\log{{\left[{B}{\left({1}\right)}{B}{\left({2}\right)}{B}{\left({3}\right)}\ldots{B}{\left({n}\right)}\right]}}}\le{14496}$ . (3 marks)

 (a) ${A}{\left({1}\right)}+{A}{\left({2}\right)}+{A}{\left({3}\right)}+\ldots+{A}{\left({n}\right)}$ $={20}+{32}+{44}+\ldots+{\left({12}{n}+{8}\right)}$ $=\dfrac{{n}}{{2}}{\left[{20}+{\left({12}{n}+{8}\right)}\right]}$ ${S}{\left({n}\right)}=\dfrac{{n}}{{2}}{\left({a}+{l}\right)}$ 1M $=\dfrac{{n}}{{2}}{\left({12}{n}+{28}\right)}$ $={6}{n}^{{2}}+{14}{n}$ 1A or equivalent  (b) ${\log{{\left[{B}{\left({1}\right)}{B}{\left({2}\right)}{B}{\left({3}\right)}\ldots{B}{\left({n}\right)}\right]}}}$ $\le{14496}$ ${\log{{B}}}{\left({1}\right)}+{\log{{B}}}{\left({2}\right)}+{\log{{B}}}{\left({3}\right)}+\ldots+{\log{{B}}}{\left({n}\right)}$ $\le{14496}$ 1M can be absorbed Note that ${\log{{B}}}{\left({k}\right)}={A}{\left({k}\right)}$ for all positive integers ${k}$ . ${A}{\left({1}\right)}+{A}{\left({2}\right)}+{A}{\left({3}\right)}+\ldots+{A}{\left({n}\right)}$ $\le{14496}$ ${6}{n}^{{2}}+{14}{n}$ $\le{14496}$ 1M ${6}{n}^{{2}}+{14}{n}-{14496}$ $\le{0}$ ${\left({n}-{48}\right)}{\left({3}{n}+{151}\right)}$ $\le{0}$ ∴  $-\dfrac{{151}}{{3}}\le{n}\le{48}$ accept ${0}\le{n}\le{48}$ Thus, the greatest value of ${n}$ is ${48}$ . 1A

 ${A}{\left({n}\right)}={12}{n}+{8}$ (a) First term $={12}{\left({n}\right)}+{8}$ $={20}$ Common difference $={A}{\left({n}+{1}\right)}-{A}{\left({n}\right)}$ $={\left[{12}{\left({n}+{1}\right)}+{8}\right]}-{\left({12}{n}+{8}\right)}$ $={12}$ ${A}{\left({1}\right)}+{A}{\left({2}\right)}+{A}{\left({3}\right)}+\ldots+{A}{\left({n}\right)}$ $=\dfrac{{n}}{{2}}{\left[{2}{\left({20}\right)}+{\left({n}-{1}\right)}{12}\right]}$ ${S}{\left({n}\right)}=\dfrac{{n}}{{2}}{\left[{2}{a}+{\left({n}-{1}\right)}{d}\right]}$ 1M $=\dfrac{{n}}{{2}}{\left({12}{n}+{28}\right)}$ $={6}{n}^{{2}}+{14}{n}$ 1A or equivalent  (b) ${\log{{\left[{B}{\left({1}\right)}{B}{\left({2}\right)}{B}{\left({3}\right)}\ldots{B}{\left({n}\right)}\right]}}}$ $\le{14496}$ ${\log{{\left({10}^{{20}}{10}^{{32}}{10}^{{44}}\right.}}}$ $\ldots$ ${10}^{{{12}{n}+{8}}}{)}$ $\le{14496}$ ${\log{{\left({10}^{{{20}+{32}+{44}+\ldots+{\left({12}{n}+{8}\right)}}}\right)}}}$ $\le{14496}$ 1M can be absorbed ${\log{{\left({10}^{{{6}{n}^{{2}}+{14}{n}}}\right)}}}$ $\le{14496}$ ${6}{n}^{{2}}+{14}{n}$ $\le{14496}$ 1M ${\left({n}-{48}\right)}{\left({3}{n}+{151}\right)}$ $\le{0}$ ∴  $-\dfrac{{151}}{{3}}\le{n}\le{48}$ accept ${0}\le{n}\le{48}$ Thus, the greatest value of ${n}$ is ${48}$ . 1A

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