NCERT Solutions · Class 11 Mathematics Sequences and Series

Convert each repeating‑digit term into a geometric series.(i) \(\displaystyle 5+55+555+\ldots\) to \(\displaystyle n\) terms.Take out the digit \(\displaystyle 5\) and write each term with nines: \[S=5\left(1+11+111+\cdots\right)=\frac{5}{9}\left(9+99+999+\cdots \text{ to } n \text{ terms}\right).\] Now \(\displaystyle 9=10-1,\;99=10^{2}-1,\;999=10^{3}-1,\ldots\), so \[S=\frac{5}{9}\Big[\left(10+10^{2}+\cdots+10^{n}\right)-n\Big].\] The bracketed sum is a G.P. with first term \(\displaystyle 10\) and ratio \(\displaystyle 10\); using \(\displaystyle S_n=\dfrac{a(r^{n}-1)}{r-1}\), \[10+10^{2}+\cdots+10^{n}=\frac{10\left(10^{n}-1\right)}{9}.\] Therefore \[S=\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]=\frac{50\left(10^{n}-1\right)}{81}-\frac{5n}{9}=\frac{5}{81}\left(10^{\,n+1}-10-9n\right).\] Check: \(\displaystyle n=2\) gives \(\displaystyle \frac{5}{81}(1000-10-18)=\frac{5\times 972}{81}=60=5+55\). ✓(ii) \(\displaystyle 0.6+0.66+0.666+\ldots\) to \(\displaystyle n\) terms.Take out the digit \(\displaystyle 6\): \[S=6\left(0.1+0.11+0.111+\cdots\right)=\frac{6}{9}\left(0.9+0.99+0.999+\cdots \text{ to } n \text{ terms}\right).\] Now \(\displaystyle 0.9=1-\dfrac{1}{10},\;0.99=1-\dfrac{1}{10^{2}},\ldots\), so \[S=\frac{2}{3}\left[n-\left(\frac{1}{10}+\frac{1}{10^{2}}+\cdots+\frac{1}{10^{n}}\right)\right].\] The inner sum is a G.P. with \(\displaystyle a=\frac{1}{10}\), \(\displaystyle r=\frac{1}{10}\): \[\frac{1}{10}+\cdots+\frac{1}{10^{n}}=\frac{\dfrac{1}{10}\left(1-\dfrac{1}{10^{n}}\right)}{1-\dfrac{1}{10}}=\frac{1}{9}\left(1-\frac{1}{10^{n}}\right).\] Therefore \[S=\frac{2}{3}\left[n-\frac{1}{9}\left(1-10^{-n}\right)\right]=\frac{2n}{3}-\frac{2}{27}\left(1-10^{-n}\right)=\frac{2}{27}\left(9n-1+10^{-n}\right).\] Check: \(\displaystyle n=2\) gives \(\displaystyle \frac{2}{27}(18-1+0.01)=\frac{2\times 17.01}{27}=1.26=0.6+0.66\). ✓Answers: (i) \(\displaystyle \dfrac{50\left(10^{n}-1\right)}{81}-\dfrac{5n}{9}\) (ii) \(\displaystyle \dfrac{2n}{3}-\dfrac{2}{27}\left(1-10^{-n}\right)\).

Write the general term of the series.The factors in the products run \(\displaystyle 2,4,6,8,\ldots\) — the even numbers. In the \(\displaystyle k^{\text{th}}\) product the first factor is \(\displaystyle 2k\) and the second is the next even number, \(\displaystyle 2k+2\): \[2\times 4,\quad 4\times 6,\quad 6\times 8,\ \ldots \Longrightarrow a_{k}=2k\,(2k+2)=4k^{2}+4k.\] Check: \(\displaystyle k=1\Rightarrow 2\times 4=8\); \(\displaystyle k=3\Rightarrow 6\times 8=48=4(9)+12\). ✓Put \(\displaystyle k=20\): \[a_{20}=2(20)\times\left(2(20)+2\right)=40\times 42=1680.\]The \(\displaystyle 20^{\text{th}}\) term is \(\displaystyle 1680\).

The interest payments form an A.P.Cost of the tractor \(\displaystyle =\) Rs \(\displaystyle 12000\); cash paid \(\displaystyle =\) Rs \(\displaystyle 6000\), so the unpaid balance is \[12000-6000=\text{Rs } 6000 .\] This is cleared in annual instalments of Rs \(\displaystyle 500\), so the number of instalments is \[\frac{6000}{500}=12 .\]Interest of \(\displaystyle 12\%\) is charged each year on the amount still unpaid. The unpaid amounts at the start of successive years are \[6000,\;5500,\;5000,\;\ldots,\;500,\] so the yearly interests are \[12\%\text{ of }6000=720,\quad 12\%\text{ of }5500=660,\quad\ldots,\quad 12\%\text{ of }500=60 .\]These \(\displaystyle 12\) numbers form an A.P. with \(\displaystyle a=720\), last term \(\displaystyle l=60\). Using \(\displaystyle S_{n}=\dfrac{n}{2}(a+l)\), \[\text{Total interest}=\frac{12}{2}\,(720+60)=6\times 780=\text{Rs } 4680 .\]Total cost \(\displaystyle =\) price of the tractor \(\displaystyle +\) total interest paid: \[12000+4680=16680 .\]The tractor costs him Rs 16680.

The interest payments form an A.P.Cost of the scooter \(\displaystyle =\) Rs \(\displaystyle 22000\); cash paid \(\displaystyle =\) Rs \(\displaystyle 4000\), so the unpaid balance is \[22000-4000=\text{Rs } 18000 .\] With annual instalments of Rs \(\displaystyle 1000\), the number of instalments is \[\frac{18000}{1000}=18 .\]Interest of \(\displaystyle 10\%\) is charged each year on the amount still unpaid. The unpaid amounts at the start of successive years are \[18000,\;17000,\;16000,\;\ldots,\;1000,\] so the yearly interests are \[1800,\;1700,\;1600,\;\ldots,\;100 .\]These \(\displaystyle 18\) numbers form an A.P. with \(\displaystyle a=1800\), last term \(\displaystyle l=100\). Using \(\displaystyle S_{n}=\dfrac{n}{2}(a+l)\), \[\text{Total interest}=\frac{18}{2}\,(1800+100)=9\times 1900=\text{Rs } 17100 .\]Total cost \(\displaystyle =\) price of the scooter \(\displaystyle +\) total interest paid: \[22000+17100=39100 .\]The scooter costs him Rs 39100.

The numbers of letters in successive sets form a G.P.The first person mails \(\displaystyle 4\) letters — that is the \(\displaystyle 1^{\text{st}}\) set. Each of those \(\displaystyle 4\) people mails \(\displaystyle 4\) letters, giving \(\displaystyle 4^{2}=16\) letters in the \(\displaystyle 2^{\text{nd}}\) set; each of those mails \(\displaystyle 4\) more, giving \(\displaystyle 4^{3}\) in the \(\displaystyle 3^{\text{rd}}\) set, and so on. So the sets contain \[4,\;4^{2},\;4^{3},\;\ldots,\;4^{8}\] letters — a G.P. with \(\displaystyle a=4\), \(\displaystyle r=4\), \(\displaystyle n=8\).Total number of letters mailed up to and including the \(\displaystyle 8^{\text{th}}\) set: \[S_{8}=\frac{a\left(r^{n}-1\right)}{r-1}=\frac{4\left(4^{8}-1\right)}{4-1}=\frac{4\,(65536-1)}{3}=\frac{4\times 65535}{3}=4\times 21845=87380 .\]Each letter costs \(\displaystyle 50\) paise \(\displaystyle =\) Rs \(\displaystyle 0.50\), so the postage is \[87380\times 0.50=\text{Rs } 43690 .\]Rs $\displaystyle 43690$ is spent on postage.

Simple interest makes the yearly amounts an A.P.Principal \(\displaystyle P=\) Rs \(\displaystyle 10000\) at \(\displaystyle 5\%\) simple interest, so the interest credited every year is the same: \[\text{S.I. per year}=\frac{10000\times 5}{100}=\text{Rs } 500 .\]Because the same Rs \(\displaystyle 500\) is added each year, the successive amounts form an A.P. with common difference \(\displaystyle d=500\). In the \(\displaystyle 1^{\text{st}}\) year the money standing is the deposit itself, Rs \(\displaystyle 10000\); in the \(\displaystyle 2^{\text{nd}}\) year it is \(\displaystyle 10500\); in the \(\displaystyle 3^{\text{rd}}\) year \(\displaystyle 11000\); and so on: \[10000,\;10500,\;11000,\;\ldots\] So the amount in the \(\displaystyle n^{\text{th}}\) year is \(\displaystyle a_{n}=10000+(n-1)\,500\).Amount in the \(\displaystyle 15^{\text{th}}\) year (it has earned interest for \(\displaystyle 14\) completed years): \[a_{15}=10000+14\times 500=10000+7000=\text{Rs } 17000 .\]Total amount after \(\displaystyle 20\) years (interest for \(\displaystyle 20\) completed years): \[10000+20\times 500=10000+10000=\text{Rs } 20000 .\]In the \(\displaystyle 15^{\text{th}}\) year the amount is Rs $\displaystyle 17000$, and after $\displaystyle 20$ years the total amount is Rs 20000.

Depreciation of \(\displaystyle 20\%\) a year is a G.P. with ratio \(\displaystyle 0.8\).If a value depreciates by \(\displaystyle 20\%\) in a year, what remains is \(\displaystyle 100\%-20\%=80\%\) of it, so each year the value is multiplied by \[1-\frac{20}{100}=\frac{4}{5}=0.8 .\]Starting from Rs \(\displaystyle 15625\), the values at the ends of successive years form a G.P.: \[15625\left(\frac{4}{5}\right),\;15625\left(\frac{4}{5}\right)^{2},\;\ldots\] so the value at the end of \(\displaystyle 5\) years is \[15625\left(\frac{4}{5}\right)^{5}=15625\times\frac{1024}{3125}=5\times 1024=5120 .\] (\(\displaystyle 15625=5^{6}\) and \(\displaystyle 5^{5}=3125\), which makes the cancellation exact.)The estimated value at the end of $\displaystyle 5$ years is Rs 5120.

Count the work in worker‑days: the daily workforce is an A.P.Let the job have been planned to finish in \(\displaystyle n\) days with all \(\displaystyle 150\) workers present. Then \[\text{total work}=150n \text{ worker-days}.\]In fact the workforce fell by \(\displaystyle 4\) each day after the first: \(\displaystyle 150\) on day $\displaystyle 1$, \(\displaystyle 146\) on day $\displaystyle 2$, \(\displaystyle 142\) on day $\displaystyle 3$, and so on — an A.P. with \(\displaystyle a=150,\;d=-4\). The job actually took \(\displaystyle n+8\) days, so the work done is the sum of \(\displaystyle n+8\) terms of this A.P.: \[S_{n+8}=\frac{n+8}{2}\Big[2(150)+(n+8-1)(-4)\Big]=\frac{n+8}{2}\Big[300-4(n+7)\Big]=(n+8)(136-2n).\]The same job was done either way, so \[(n+8)(136-2n)=150n.\] Expanding, \[136n-2n^{2}+1088-16n=150n\;\Longrightarrow\; -2n^{2}-30n+1088=0\;\Longrightarrow\; n^{2}+15n-544=0.\] Then \[n=\frac{-15\pm\sqrt{225+2176}}{2}=\frac{-15\pm\sqrt{2401}}{2}=\frac{-15\pm 49}{2},\] and since \(\displaystyle n>0\), \(\displaystyle n=17\).So the work was planned for \(\displaystyle 17\) days and actually took \(\displaystyle 17+8=25\) days.Check: over \(\displaystyle 25\) days the workforce runs \(\displaystyle 150,146,\ldots,150-4(24)=54\), and \[S_{25}=\frac{25}{2}(150+54)=25\times 102=2550=150\times 17. \checkmark\]The work was completed in $\displaystyle 25$ days.