Chapter 8: Sequences and Series

Ans: Let \[{{S}_{n}}=5+55+555...\] to \[n\] terms.

\[=\frac{5}{9}\](9+99+999+... to n terms.)

\[=\frac{5}{9}(( 10-1 )+( {{10}^{2}}-1 )+( {{10}^{3}}-1)+...\]to n terms)

\[=\frac{5}{9}((10+{{10}^{2}}+{{10}^{3}}...\]to n terms)-(1+1+ to n terms))

\[=\frac{5}{9}\left[ \frac{10\left( {{10}^{n}}-1 \right)}{10-1}-n \right]\]

\[=\frac{5}{9}\left[ \frac{10\left( {{10}^{n}}-1 \right)}{9}-n \right]\]

\[=\frac{50}{81}\left( {{10}^{n}}-1 \right)-\frac{5n}{9}\]

Therefore, the sum of \[n\] terms of the given series is \[\frac{50}{81}\left( {{10}^{n}}-1 \right)-\frac{5n}{9}\] .

Ans: Let \[{{S}_{n}}=0.6+0.66+0.666+\] to \[n\] terms.

\[=6\] (0.1+0.11+0.111+... to \[n\] terms)

\[=\frac{6}{9}\] (0.9+0.99+0.999+... to \[n\] terms)

\[=\frac{6}{9} (\left( 1-\frac{1}{10} \right)+\left( 1-\frac{1}{{{10}^{2}}} \right)+\left( 1-\frac{1}{{{10}^{3}}} \right))+... \]to n terms

\[=\frac{2}{3}\]((\[1+1+...\] to \[n\] terms)\[-\] \[\frac{1}{10}\] (\[1+\frac{1}{10}+\frac{1}{{{10}^{2}}}\] to \[n\] terms))

\[=\frac{2}{3}( n-\frac{1}{10}\left( \frac{1-{{\left( \frac{1}{10} \right)}^{n}}}{1-\frac{1}{10}} \right) )\]

\[=\frac{2}{3}n-\frac{2}{30}\times \frac{10}{9}\left( 1-{{10}^{-n}} \right)\]

\[=\frac{2}{3}n-\frac{2}{27}\left( 1-{{10}^{-n}} \right)\]

Therefore, the sum of \[n\] terms of the given series is \[\frac{2}{3}n-\frac{2}{27}\left( 1-{{10}^{-n}} \right)\] .

Ans: \[2\times 4+4\times 6+6\times 8+...+n\] is the given series,

Therefore the \[{{n}^{th}}\] term \[{{a}_{n}}=2n\times \left( 2n+2 \right)=4{{n}^{2}}+4n\]

Then,

\[{{a}_{20}}=4{{\left( 20 \right)}^{2}}+4\left( 20 \right)\]

\[=4\left( 400 \right)+80\]

\[=1600+80\]

\[=1680\]

Therefore, \[1680\] is the \[{{20}^{th}}\] term of the series. 

Ans: It is given that Rs.\[6000\] is paid in cash by the farmer.

Therefore, the unpaid amount is given by

Rs.\[12000-\] Rs.\[6000=\]Rs.\[6000\]

According to the conditions given in the question, the interest to be paid annually by the farmer is 

\[12%\] of \[6000\] , \[12%\] of \[5500\] , \[12%\] of \[5000...12%\] of \[500\]

Therefore, the total interest to be paid by the farmer

\[=12%\] of \[6000+12%\] of \[5500+12%\] of \[5000+...+12%\] of \[500\]

\[=12%\] of \[\left( 6000+5500+5000+...+500 \right)\]

\[=12%\] of \[\left( 500+1000+1500+...+6000 \right)\]

With both the first term and common difference equal to \[500\], the series \[500,1000,1500...6000\] is an A.P.

Let \[n\] be the number of terms of the A.P. 

Therefore,

\[6000=500+\left( n-1 \right)500\]

\[\Rightarrow 1+\left( n-1 \right)=12\]

\[\Rightarrow n=12\]

Therefore, the sum of the given A.P.

\[=\frac{12}{2}\left[ 2\left( 500 \right)+\left( 12-1 \right)\left( 500 \right) \right]\]

\[=6\left[ 1000+5500 \right]\]

\[=6\left( 6500 \right)\]

\[=39000\]

Therefore, the total interest to be paid by the farmer

\[=12%\] of \[\left( 500+1000+1500+...+6000 \right)\]

\[=12%\] of Rs.\[39000\]

\[=\] Rs.\[4680\] 

Therefore, the total cost of a tractor

\[=\](Rs.\[12000+\]Rs.\[4680\])

\[=\]Rs.\[16680\]

Therefore, the total cost of the tractor is Rs.\[16680\].

Ans: It is given that for Rs.\[22000\] Shamshad Ali buys a scooter and Rs.\[4000\] is paid in cash.

Therefore, the unpaid amount is given by

Rs.\[22000-\] Rs.\[4000=\]Rs.\[18000\]

According to the conditions given in the question, the interest to be paid annually

is 

\[10%\] of \[18000\] , \[10%\] of \[17000\] , \[10%\] of \[16000...10%\] of \[1000\]

Therefore, the total interest to be paid by the farmer

\[=10%\] of \[18000+10%\] of \[17000+10%\] of \[16000+...+10%\] of \[1000\]

\[=10%\] of \[\left( 18000+17000+16000+...+1000 \right)\]

\[=10%\] of \[\left( 1000+2000+3000+...+18000 \right)\]

With both the first term and common difference equal to \[1000\], the series \[1000,2000,3000...18000\] is an A.P.

Let \[n\] be the number of terms of the A.P. 

Therefore,

\[18000=1000+\left( n-1 \right)1000\]

\[\Rightarrow 1+\left( n-1 \right)=18\]

\[\Rightarrow n=18\]

Therefore, the sum of the given A.P.

\[=\frac{18}{2}\left[ 2\left( 1000 \right)+\left( 18-1 \right)\left( 1000 \right) \right]\]

\[=9\left[ 2000+17000 \right]\]

\[=9\left( 19000 \right)\]

\[=171000\]

Therefore, the total interest to be paid

\[=10%\] of \[\left( 18000+17000+16000+...+1000 \right)\]

\[=10%\] of  Rs.\[171000\]

\[=\] Rs.\[17100\] 

Therefore, the total cost of the scooter

\[=\](Rs.\[22000+\]Rs.\[17100\])

\[=\]Rs.\[39100\]

Therefore, the total cost of the scooter is Rs.\[39100\] .

Ans: \[4,{{4}^{2}},{{...4}^{8}}\] is the number of letters mailed and it forms a G.P.

The first term \[a=4\] , the common ratio \[r=4\] and the number of terms \[n=8\] of the G.P.

We know that the sum of \[n\] terms of a G.P. is 

\[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\]

Therefore,

\[{{S}_{8}}=\frac{4\left( {{4}^{8}}-1 \right)}{4-1}\]

\[=\frac{4\left( 65536-1 \right)}{3}\]

\[=\frac{4\left( 65535 \right)}{3}\]

\[=4\left( 21845 \right)\]

\[=87380\]

\[50\] paisa is the cost of mailing one letter.

Therefore,

Cost of mailing \[87380\] letters \[=\] Rs.\[87380\times \frac{50}{100}\] \[=\] Rs.\[43690\]

Therefore, Rs.\[43690\] is the amount spent when \[{{8}^{th}}\] set of letter is mailed.

Ans: Rs.\[10000\] is deposited by the man in a bank at the rate of \[5%\] simple interest annually

\[=\frac{5}{100}\times \]Rs.\[10000=\]Rs.\[500\]

Therefore,

\[10000+500+500+...+500\] is the interest in \[{{15}^{th}}\] year. (\[500\] is \[14\] added times)

Therefore, the amount in \[{{15}^{th}}\] year 

\[=\]Rs.\[10000+14\times \]Rs.\[500\]

\[=\]Rs.\[10000+\]Rs.\[7000\]

\[=\]Rs.\[17000\]

Rs.\[10000+500+500+...+500\] is the amount after \[20\] years. (\[500\] is \[20\] added times)

Therefore, the amount after \[20\] years

\[=\]Rs.\[10000+20\times \]Rs.\[500\]

\[=\]Rs.\[10000+\]Rs.\[10000\]

\[=\]Rs.\[20000\]

The total amount after \[20\] years is Rs.\[20000\].

Ans: The cost of the machine is Rs.\[15625\].

Every year machine depreciates by \[20%\].

Therefore, \[80%\] of the original cost,i.e., \[\frac{4}{5}\] of the original cost is its value after every year.

Therefore, the value at the end of \[5\] years 

\[=15626\times \frac{4}{5}\times \frac{4}{5}\times ...\times \frac{4}{5}\]

\[=5\times 1024\]

\[=5120\]

Therefore, Rs.\[5120\] is the value of the machine at the end of \[5\] years.

Ans: Let the number of days in which \[150\] workers finish the work be \[x\].

According to the conditions given in the question,

\[150x=150+146+142+...\left( x+8 \right)\]terms

With first term \[a=146\], common difference \[d=-4\] and number of turns as \[\left( x+8 \right)\] , the series \[150+146+142+...\left( x+8 \right)\]terms is an A.P. 

\[\Rightarrow 150x=\frac{\left( x+8 \right)}{2}\left[ 2\left( 150 \right)+\left( x+8-1 \right)\left( -4 \right) \right]\]

\[\Rightarrow 150x=\left( x+8 \right)\left[ 150+\left( x+7 \right)\left( -2 \right) \right]\]

\[\Rightarrow 150x=\left( x+8 \right)\left( 150-2x-14 \right)\]

\[\Rightarrow 150x=\left( x+8 \right)\left( 136-2x \right)\]

\[\Rightarrow 75x=\left( x+8 \right)\left( 68-x \right)\]

\[\Rightarrow 75x=68x-{{x}^{2}}+544-8x\]

\[\Rightarrow {{x}^{2}}+75x-60x-544=0\]

\[\Rightarrow {{x}^{2}}+15x-544=0\]

\[\Rightarrow {{x}^{2}}+32x-17x-544=0\]

\[\Rightarrow x\left( x+32 \right)-17\left( x+32 \right)=0\]

\[\Rightarrow \left( x-17 \right)\left( x+32 \right)=0\]

\[\Rightarrow x=17\] or \[x=-32\]

We know that \[x\] cannot be negative.

So, \[x=17\].

Therefore, \[17\] is the number of days in which the work was completed. Then the required number of days \[=\left( 17+8 \right)=25\] .