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The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?

Let she fixes 13 flags on one side and 13 on other side of the middle flag.

distance covered for fixing first flag and return is 2m + 2m = 4m

distance covered for fixing second flag and return is 4m + 4m = 8m

distance covered for fixing third flag and return is 6m + 6m = 12m

Similarly for thirteen flags is

4m + 8m + 12m + …….

Here, a = 4

d = 8 - 4 = 4

n = 13

\\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ \mathrm{S}_{13}=\frac{13}{2}[2(4)+(13-1) 4] \\ \frac{13}{2}[56]=364

Total distance covered = 364 × 2 = 728 m         (for both side )

distance, only to carry the flag = 364.

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Jaspal Singh repays his total loan of Rs 118000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, what amount will be paid by him in the 30thinstalment? What amount of loan does he still have to pay after the 30th instalment?

Total loan = 118000

                        First installment (a) = 1000

                        Installment increased = 100

                        Hence d = 100

                        30\textsuperscript{th} $installment = a\textsubscript{30} = a + 29d \\

= 1000 + 29(100)

= 1000 + 2900

= 3900

Amount of loan paid after 30 installments

= Total loan - 30th installments is total loan

\\118000 - \frac{30}{2}\left[ 2\times 1000+\left( 30-1 \right)\times 100 \right] \\ 118000 - 15 [2000 + 2900]\\ 118000 - 15 [4900]\\ 118000 - 73500 = 44500\\

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Solve the equation- 4 + (-1) + 2 +...+ x = 437

\\$Given series :$\\ -4 + (-1) + 2 + $ \ldots $ + x = 437\\ $Here a $= - 4\\ d = - 1 - (-4)\\ -1 + 4 = + 3\\ S\textsubscript{n} = 437 $ (given)$\\ $We know that $ {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\437=\frac{n}{2}[2(-4)+(n-1) 3]\\ 874 = n[- 8 + 3n - 3]\\ 874 = 3n\textsuperscript{2} - 11n\\ 3n\textsuperscript{2} - 11n - 874 = 0\\ 3n\textsuperscript{2} - 57n + 46n - 874 = 0\\ 3n (n - 19) + 46 (n - 19) = 0 \\ (n - 19) (3n + 46) = 0\\ n - 19 = 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2n + 16 = 0\\ n = 19 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: n = - 46/3\\ $Hence $n = 19 \\ a\textsubscript{n} = x \\ a + (n - 1)d = x $ (using a\textsubscript{n }$= a + (n-1)d )\\ -4 + (19 - 1)3 = x\\ -4 + (18)3 = x\\ -4 + 54 = x\\ x = 50\\

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Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to

 \frac{(a+c)(b+c-2 a)}{2(b-a)}

Here

 a\textsubscript{1} = a, a\textsubscript{2} = b\\

\\ d = a\textsubscript{2} - a\textsubscript{1}\\ = b - a \\ a\textsubscript{n} = c \: \: \: \: \: \: \: \: \: \: \: \: \because a\textsubscript{n} =a + (n - 1)d \\ a\textsubscript{1} + (n - 1).d = c\: \: \: \: \: $ Put a\textsubscript{1}$ = a , d= b - a \\ (n - 1) (b - a) = c - a

\\ \mathrm{n}-1=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{n}=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{b}-\mathrm{a}}+1 \\ \mathrm{n}=\frac{\mathrm{c}-\mathrm{a}+\mathrm{b}-\mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{n}=\frac{\mathrm{c}+\mathrm{b}-2 \mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2 \mathrm{a}_{1}+(\mathrm{n}-1) \mathrm{d}\right] \\ =\frac{\mathrm{c}+\mathrm{b}-2 \mathrm{a}}{2(\mathrm{~b}-\mathrm{a})}[\mathrm{a}+\mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ =\frac{(\mathrm{c}+\mathrm{b}-2 \mathrm{a})}{2(\mathrm{~b}-\mathrm{a})}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right] \quad\left[\because \mathrm{a}+(\mathrm{n}-1) \mathrm{d}=\mathrm{a}_{\mathrm{n}}\right] \\ \frac{(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{c}-2 \mathrm{a})}{2(\mathrm{~b}-\mathrm{a})} \quad \quad\left[\because \mathrm{a}_{\mathrm{n}}=\mathrm{c}\right]

Hence proved.

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The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5thterm to the 21stterm, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.

According to question

\\\frac{{{a}_{11}}}{{{a}_{18}}}=\frac{2}{3} \Rightarrow \frac{a+10d}{a+17d}=\frac{2}{3} (Using a\textsubscript{n} = a + (n - 1).d )\\ \\ 3a + 30d = 2a + 34d\\ 3a - 2a + 30d - 34d = 0\\ a - 4d = 0 a = 4d $ \ldots $ (1)\\
\\$Ratio of 5\textsuperscript{th} and 21th term is $\\ \frac{{{a}_{5}}}{{{a}_{21}}}=\frac{a+4d}{a+20d} $ \ldots $ (2)\\ $Put a = 4d in (2) we get $ =\frac{4d+4d}{4d+20d}=\frac{8d}{24d} \\

\ \frac{\mathrm{a}_{5}}{\mathrm{a}_{21}}=\frac{1}{3}$ \\Ratio of $\mathrm{S}_{5}$ to $\mathrm{S}_{21}$ is \\$\frac{\mathrm{S}_{5}}{\mathrm{~S}_{21}}=\frac{\frac{5}{2}[2 \mathrm{a}+4 \mathrm{~d}]}{\frac{21}{2}[2 \mathrm{a}+20 \mathrm{~d}]} \quad\left[\because \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]\right]$ \\$\frac{\mathrm{S}_{5}}{\mathrm{~S}_{21}}=\frac{5[2 \mathrm{a}+4 \mathrm{~d}]}{21[2 \mathrm{a}+20 \mathrm{~d}]}$ \\Put $a=4 d$ in equation (3) we get \\$\frac{\mathrm{S}_{5}}{\mathrm{~S}_{21}}=\frac{5[2(4 \mathrm{~d})+4 \mathrm{~d}]}{21[2(4 \mathrm{~d})+20 \mathrm{~d}]}=\frac{5[8 \mathrm{~d}+4 \mathrm{~d}]}{21[8 \mathrm{~d}+20 \mathrm{~d}]}=\frac{5[12 \mathrm{~d}]}{21[28 \mathrm{~d}]}=\frac{60 \mathrm{~d}}{588 \mathrm{~d}}$ \\$\frac{\mathrm{S}_{5}}{\mathrm{~S}_{21}}=\frac{5}{49}$

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i)Find the sum of the integers between 100 and 200 that are divisible by 9

ii)Find the sum of the integers between 100 and 200 that are not divisible by 9

i)

Numbers between 100 and 200 which is divisible by 9 are 108, 117, 126, ... 198. 

 Here a = 108, d = 117- 108 = 9 

\\ a\textsubscript{n} = 198\\ a + (n - 1) \times d = 198\\ 108 + (n - 1) \times 9 = 198\\ (n - 1) \times 9 = 198 - 108\\ \left( n-1 \right)=\frac{90}{9} \\

\\n = 10 + 1 = 11 \\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ \text { Put } \mathrm{n}=11 \\ \begin{aligned} \mathrm{S}_{11} &=\frac{11}{2}[2(108)+(11-1) 9] \\ &=\frac{11}{2}[216+90] \\ &=\frac{11}{2} \times 306 \\ &=11 \times 153 \\ \mathrm{~S}_{11} &=1683 \end{aligned}

ii)

Numbers between 100 and 200 are 101, 102, 103,   .. , 199 

 Here a = 101, d = 102 - 101 = 1  

\\a\textsubscript{n} = 199\\ a + (n - 1).d = 199 ( Using a\textsubscript{n} = a + (n - 1)d )\\ 101 + (n - 1)1 = 199\\ (n - 1) = 199 - 101\\ n = 98 + 1\\ n = 99\\ {{S}_{99}}=\frac{99}{2}\left[ 2\times 101+\left( 99-1 \right)\left( 1 \right) \right] \\
\begin{aligned} & =\frac{99}{2}\left[ 202+98 \right] \ & =\frac{99}{2}\left[ 300 \right] \ \end{aligned} \\

\\=99[150]\\ = 14850\\

Numbers between 100 and 200 which is divisible by 9 are 108, 117, 126,  ...198 

 Here b = 108,

 d\textsubscript{1} = 117- 108 = 9\\

\\ b\textsubscript{N} = 198\\ b + (N - 1) \times d\textsubscript{1 } = 198\\ 108 + (N - 1) \times 9 = 198\\ (N - 1) \times 9 = 198 - 108\\ \left( N-1 \right)=\frac{90}{9} \\

\\N= 10 + 1 = 11\\ \mathrm{S}_{\mathrm{N}}=\frac{\mathrm{N}}{2}\left[2 \mathrm{~b}+(\mathrm{N}-1) \mathrm{d}_{1}\right] \\ \text { Put } \mathrm{N}=11 \\ \begin{aligned} \mathrm{S}_{11} &=\frac{11}{2}[2(108)+(11-1) 9] \\ &=\frac{11}{2}[216+90] \\ &=\frac{11}{2} \times 306 \\ &=11 \times 153 \\ \mathrm{~S}_{11} &=1683 \end{aligned}

 the sum of the integers between 100 and 200 that is not divisible by 9   

=Sum of all numbers - the sum of numbers which is divisible by 9

\\ = 14850 - 1683 \\ = 13167\\

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An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.

Total terms = 37,  middle term =

\frac{37+1}{2}=19 \\

Three middle most terms, 18th , 19thand 20th, given

 a\textsubscript{18} + a\textsubscript{19} + a\textsubscript{20} = 225 \\

\\a + 17d + a + 18d + a + 19d = 225 $ ( Using a\textsubscript{n} $= a + (n - 1)d ) \\ 3a + 54d = 225\\ $dividing by 3 we get\\$ a + 18d = 75\\ a = 75 - 18d $ \ldots $ (1)\\ $According to question\\$ a\textsubscript{35} + a\textsubscript{36} + a\textsubscript{37} = 429 ( $Using a\textsubscript{n} $= a + (n - 1)d ) \\ a + 34d + a+ 35d + a + 36d = 429\\ 3a + 105d = 429\\ $Dividing by three we get\\$ a + 35d = 143\\ $Put value from equation (1) we get\\$ 75 - 18d + 35d = 143, \\ 17d = 143 - 75\\ d = 68, \\ \begin{aligned} & d=\frac{68}{17} \ & d=4 \ \end{aligned} \\ $Put d = 4 in (1) we get\\$ a = 75 - 72 = 3\\ a + d = 3 + 4 = 7\\ a + 2d = 3 + 2(4) = 11\\ $Required AP is$ 3, 7, 11, $ \ldots $ $ \ldots $ .\\

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The eighth term of an AP is half its second term and the eleventh term exceeds

one third of its fourth term by 1. Find the 15thterm.

According to question

$\\ \\a\textsubscript{8 }= a\textsubscript{2}/2 \\ 2a\textsubscript{8} = a\textsubscript{2}\\ 2(a + 7d) = a + d (Using a\textsubscript{n} = a + (n - 1)d ) \\ 2a + 14d - a - d = 0\\ a + 13d = 0 $ \ldots $ (1)\\ 3a\textsubscript{11} = a\textsubscript{4} + 3\\ 3(a + 10d) = a + 3d + 3 (Using a\textsubscript{n} = a + (n - 1)d ) \\ 3a + 30d - a - 3d = 3\\ 2a + 27d = 3 $ \ldots $ (2)\\ Apply: eq.(2) - 2$ \times $ eq.(1) we get\\ 2a + 27d - 2a - 26d = 3\\ d = 3\\ Put d = 3 in (1) we get\\ a +13(3) = 0\\ a = - 39\\ Then a\textsubscript{15} = a + 14d (Using a\textsubscript{n} = a + (n - 1)d ) \\ = - 39 + 14(3)\\ = -39 + 42 \\ = 3\\

 

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i)Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.

ii)Find the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5 .

iii)Find the sum of those integers from 1 to 500 which are multiples of 2 or 5.

i)

\\$The numbers which are multiples of 2 as well as 5 i.e. multiple of 10.\\ 10, 20, 30 $ \ldots $ $ \ldots $ . 490.\\ Here first term (a) $= 10\\ $ Common difference (d) $= 20 - 10 = 10\\ a\textsubscript{n} = 490\\ a + (n - 1).d = 490\\ 10 + (n - 1)10 = 490\\ (n - 1)10 = 490 - 10\\ \left( n-1 \right)=\frac{480}{10} \\ n = 48 + 1\\ n = 49\\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]=\frac{49}{2}[10+490]=\frac{49}{2}[500] \\ = 49[250] = 12250\\

ii)

\\$The numbers which are multiples of 2 as well as 5 is the multiple of 10.\\ 10, 20, 30, $ \ldots $ .., 500\\ Here first term (a) = 10\\ Common difference (d) = 20 - 10 = 10\\$ a\textsubscript{n} = 500 \because a\textsubscript{n}=a + (n - 1)d \\ a + (n - 1)d = 500\\ 10 + (n - 1)10 = 500\\ (n - 1) 10 = 500 - 10\\ n-1=\frac{490}{10} \\ n = 49 + 1\\ n = 50\\ {{S}_{n}}=\frac{n}{2}\left[ a+{{a}_{n}} \right] put n =50\\ 25[510] = 12750\\

iii)

\\$The numbers which are multiples of 2 are 2, 4, 6, 8, $ \ldots $ $ \ldots $ , 500\\ Here first term (a\textsubscript{1}) = 2, common difference$(d\textsubscript{1}) = 4 - 2, {{n}_{_{1}}}=\frac{500}{2}=250 \\ {{a}_{{{n}_{_{1}}}}} = 500\\ \mathrm{S}_{\mathrm{n}_{1}}=\frac{\mathrm{n}_{1}}{2}\left[\mathrm{a}_{1}+\mathrm{a}_{\mathrm{n}_{1}}\right] \\ \mathrm{S}_{\mathrm{n}_{1}}=\frac{250}{2}[2+500] \\ \mathrm{S}_{\mathrm{n}_{1}}=125[502] \\ \mathrm{S}_{\mathrm{n}_{1}}=62750\\ $The numbers which are multiples of 5 are 5, 10, 15, $ \ldots $ .., 500\\ Here a\textsubscript{2} = 5, d\textsubscript{2} = 10 - 5 = 5\\$ a _{{{n}_{2}}} = 500\\ {{S}_{{{n}_{2}}}}=\frac{{{n}_{2}}}{2}\left[ {{a}_{2}}+{{a}_{{{n}_{2}}}} \right] \Rightarrow {{S}_{{{n}_{2}}}}=\frac{100}{2}\left[ 5+500 \right]=50\left( 505 \right)=25250 \\ $The numbers which are multiples of 10 are 10, 20, 30, $ \ldots $ .., 50\\ Here$ a\textsubscript{3} = 10, d\textsubscript{3} = 20 - 10 = 10\\ {{n}_{3}}=\frac{500}{10}=50 \\

\\ \mathrm{S}_{\mathrm{n}_{3}}=\frac{\mathrm{n}_{3}}{2}\left[\mathrm{a}_{3}+\mathrm{a}_{\mathrm{n}_{3}}\right] \\\\ \mathrm{S}_{\mathrm{n}_{3}}=\frac{50}{2}[10+500]=25[510]=12750

The sum of integers from 1 to 500 which are multiples of 2 or 5 = sum of integers which are multiples of 2 + sum of integer which are multiple of 5 - the sum of integer which is multiple of 10 

(we subtract the sum of integers multiple of 10 because those integers are multiples of both 2 and 5 )

=\mathrm{S}_{\mathrm{r}_{1}}+\mathrm{S}_{\mathrm{n}_{2}}-\mathrm{S}_{\mathrm{n}_{3}}=62750+25250-12750

= 88000 - 12750 = 75250 

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The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.

\\S\textsubscript{5} + S\textsubscript{7} = 167 $ \ldots $ (1)\\ S\textsubscript{10} = 235 $ \ldots $ (2)\\

 We know that

\\{{S}_{n}}=\frac{n}{2}\left( 2a+\left( n-1 \right)d \right) \\ $Put n $= 5, {{S}_{5}}=\frac{5}{2}\left( 2a+\left( 5-1 \right)d \right) \\ {{S}_{5}}=\frac{5}{2}\left( 2a+4d \right) \\ $Put n $= 7, {{S}_{7}}=\frac{7}{2}\left( 2a+\left( 7-1 \right)d \right) \\ {{S}_{7}}=\frac{7}{2}\left( 2a+6d \right) \\ $Put n $= 10, {{S}_{10}}=\frac{10}{2}\left( 2a+\left( 10-1 \right)d \right) \\ S\textsubscript{10} = 5(2a + 9d)\\ $Now put the values of S\textsubscript{5}, S\textsubscript{7} and S\textsubscript{7} in equation (1) and (2) we get\\ $ \frac{5}{2}\left( 2a+4d \right)+\frac{7}{2}\left( 2a+6d \right)=167 \\ 5a + 10d + 7a + 21d = 167\\ 12a + 31d = 167 $ \ldots $ (3)\\ 5(2a + 9d) = 235\\ 2a+9d=\frac{235}{5} \\ 2a + 9d = 47 $ \ldots $ (4)\\ $Solving equation (3) and (4)\\$ 12a + 31d = 167 $ \ldots $ (3)\\ 12a + 54d = 282 $ \{ $ $Multiply (4) by 6$$ \} $ \\
\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ -23d = - 115\\ d=\frac{-115}{-23} \\ d = 5\\ Put d = 5 in (3) we get\\ 12a + 31(5) = 167 \\ 12a = 167 - 155 \\ 12a = 12\\ a=\frac{12}{12}=1 \\ = 10 [2 + 95]\\ S\textsubscript{20} = 970\\

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