ARITHMETIC PROGRESSIONS-Exercise 5.3

Q1. Find the sum of the following APs:

1. 2, 7, 12, . . ., to 10 terms.
2. –37, –33, –29, . . ., to 12 terms.
3. 0.6, 1.7, 2.8, . . ., to 100 terms.
4. \(\frac{1}{15},\ \frac{1}{12},\ \frac{1}{10},\ \ldots\ \) to 11 terms

Solution (i):

AP: 2,7, 12, .... to to terms Sum of AP $$\boxed{S_{n}=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] }$$ $$ \begin{aligned} a&=2\\ d&=7-2\\ &=5\\ n&=10 \end{aligned} $$

Substituting values in the Formula

$$\scriptsize\begin{aligned}S_{10}&=\dfrac{10}{2}\left[ 2\cdot 2+\left( 10-1\right) 5\right] \\ &=5\left[ 4+9\times 5\right] \\ &=5\left( 4+45\right) \\ &=5\times 49\\ &=245\end{aligned}$$

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Solution: (ii)

AP: \(-37,\ -33,\ -29,\ ....\) to 12 terms

Sum of the series is given by

$$\begin{aligned} S_{n}=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right]\end{aligned}$$ Here, $$\begin{aligned} a&=-37\\ d&=-33-\left( -37\right) \\ &=4\\ n&=12\\ \end{aligned}$$

Substituting given values in the Formula

$$\scriptsize\begin{aligned}S_{12}&=\dfrac{12}{2}\left[ -37\times 2+\left( 12-1\right) 4\right] \\ &=6\left[ -74+11\times 4\right] \\ &=6\left( -74+44\right) \\ &=6\times \left( -30\right) \\ &=-180\end{aligned}$$

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Solution: (iii)

AP: 0.6, 1.7, 2.8... to 100 terms Sum of AP series is given by formula $$S_{n}=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] $$ Here, $$\begin{aligned} a&=0.6\\ d&=1.7-0.6\\ &=1.1\\ n&=100\end{aligned}$$ Substituting values in Formula $$\scriptsize\begin{aligned}S_{100}&=\dfrac{100}{2}\left[ 2\times 0.6+\left( 100-1\right) 1.1\right] \\ &=50\left[ 1.2+99\times 1.1\right] \\ &=50\left[ 1.2+108.9\right] \\ &=50\left( 110.1\right) \\ &=5505\end{aligned}$$

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Solution: iv

AP: \(\dfrac{1}{15},\ \dfrac{1}{12},\ \dfrac{1}{10},\ ....\) to 11 terms

Sum of the series is given by formula $$S_{n}=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] $$ Here, $$\begin{aligned}a&=\dfrac{1}{15}\\ d&=\dfrac{1}{12}-\dfrac{1}{15}\\ &=\dfrac{5-4}{60}\\ &=\dfrac{1}{60}\\ n&=11\end{aligned}$$ Substituting values in Formula $$\scriptsize\begin{aligned}S_{11}&=\dfrac{11}{2}\left[ 2\cdot \dfrac{1}{15}+\left( 11-1\right) \dfrac{1}{60}\right] \\ &=\dfrac{11}{2}\left[ \dfrac{2}{15}+\dfrac{10}{60}\right] \\ &=\dfrac{11}{2}\left[ \dfrac{2}{15}+\dfrac{1}{6}\right] \\ &=\dfrac{11}{2}\left[ \dfrac{4+5}{30}\right] \\ &=\dfrac{11}{2}\times \dfrac{9}{30}\\ &=\dfrac{11}{2}\times \dfrac{3}{10}\\ &=\dfrac{33}{20}\end{aligned}$$

Q2. Find the sums given below :

1. \(7 + 10\frac{1}{2} + 14 + . . . + 84\)
2. 34 + 32 + 30 + . . . + 10
3. –5 + (–8) + (–11) + . . . + (–230)

Solution: i

\(\text{AP: }7+10\dfrac{1}{2}+14+\ldots +84\) To find Sum of series, calculate number of terms $$\begin{aligned} a&=7\\ d&=10\dfrac{1}{2}-7\\ &=\dfrac{21}{2}-7\\ &=\dfrac{21-14}{2}\\ &=\dfrac{7}{2}\\ a_{n}&=84\\\\ a_{n}&=a_{1}+\left( n-1\right) d\\ 84&=7+\left( n-1\right) \dfrac{7}{2}\\ 84-7&=\dfrac{7\left( n-1\right) }{2}\\ 77\times 2&=7\left( n-1\right) \\ n-1&=\dfrac{77\times 2}{7}\\ &=22\\ n&=22+1\\ &=23\end{aligned}$$ Sum of the series is given by formula $$S_{n}=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] $$ Substituting value in the Formula $$\scriptsize\begin{aligned}S_{23}&=\dfrac{23}{2}\left[ 2\times 7+\left( 23-1\right) \dfrac{7}{2}\right] \\ &=\dfrac{23}{2}\left[ 14+\dfrac{22\times 7}{2}\right] \\ &=\dfrac{23}{2}\left( 14+77\right) \\ &=\dfrac{23}{2}\times 91\\ &=1046\dfrac{1}{2}\end{aligned}$$

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Solution: ii

\(\text{AP: }34+32+30+\ldots +10\)

To calculate the sum of series, find last term

First term \(a_{1}=34\)
Last term \(a_{n}=10\)
Common Diffrence \(d=32-34 =-2\)

$$\begin{aligned} a_{n}&=a_{1}+\left( n-1\right) d\\ 10&=34+\left( n-1\right) \left( -2\right) \\ -24&=\left( n-1\right) \left( -2\right) \\ n-1&=\dfrac{24}{2}\\ &=12\\ n&=12+1\\ &=13\end{aligned}$$ Sum of the series is given by $$S_{n}=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] $$ Subtituting values $$\scriptsize\begin{aligned}S_{13}&=\dfrac{13}{2}\left[ 2\times 34+\left( 13-1\right) \left( -2\right) \right] \\ &=\dfrac{13}{2}\left[ 68+12\times \left( -2\right) \right] \\ &=\dfrac{13}{2}\left( 68-24\right) \\ &=\dfrac{13}{2}\left( 44\right) \\ &=13\times 22\\ &=286\end{aligned}$$

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Solution: iii

$$\text{AP: }-5+\left( -8\right) +\left( -11\right) +\ldots +\left( -230\right) $$

To calculate sum of the series, lets find total number of terms in series

First term \({1}=-5\)
Last term \({n}=-230\)
Common difference \(d=-8-\left( -5\right)=-3\)

$$\begin{aligned} a_{n}&=a_{1}+\left( n-1\right) d\\ -230&=-5+\left( n-1\right) \left( -3\right) \\ -225&=-3\left( n-1\right) \\ \Rightarrow n-1&=\dfrac{225}{3}\\ n-1&=75\\ n&=75+1\\ n&=76\end{aligned}$$ Sum of series is given by Formula $$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ &=\dfrac{76}{2}\left[ 2\times \left( -5\right) +\left( 76-1\right) \times \left( -3\right) \right] \\ &=38\left[ -10+75\times \left( -3\right) \right] \\ &=38\left[ -10-225\right] \\ &=38\times \left( -235\right) \\ &=-8930\end{aligned}$$

In An AP

1. given \(a = 5, d = 3, a_n= 50,\) find \(n\) and \(S_n\)
2. \(\text{given }a = 7, a_{13} = 35,\text{ find }d\text{ and }S_{13}.\)
3. \(\text{given } a_{12}= 37, d = 3,\text{ find }a\text{ and }S_{12}.\)
4. \(\text{given } a_3 = 15, S_{10}= 125,\text{ find }d\text{ and }a_{10}.\)
5. \(\text{given } d = 5, S_9= 75,\text{ find }a\text{ and }a_9.\)
6. \(\text{given } a = 2, d = 8, S_n = 90,\text{ find }n\text{ and }a_n .\)
7. \(\text{given } a = 8, a_n= 62, S_n= 210,\text{ find }n\text{ and }d.\)
8. \(\text{given } a_n= 4, d = 2, S_n= –14,\text{ find }n\text{ and }a.\)
9. \(\text{given } a = 3, n = 8, S = 192,\text{ find }d.\)
10. \(\text{given } l = 28, S = 144,\text{ and there are total 9 terms. Find } a.\)

Solution: i

Given \(a = 5, d = 3, a_n= 50,\) find \(n\) and \(S_n\)

To find \(n\) and \(S_n\)

$$\begin{aligned}a_{n}&=a_{1}+\left( n-1\right) d\\ 50&=5+\left( n-1\right) 3\\ 45&=3\left( n-1\right) \\ n-1&=\dfrac{45}{3}\\ &=15\\ n&=16\end{aligned}$$ Sum of the series is given by $$S_{n}=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right]$$ Substituting given values $$\scriptsize\begin{aligned}S_{16}&=\dfrac{16}{2}\left[ 2\times 5+\left( 16-1\right) \times 3\right] \\ &=8\left[ 10+15\times 3\right] \\ &=8\left( 10+45\right) \\ &=8\times 55\\ &=440\\\\ \end{aligned}$$ \[\boxed{n=16,\ S_{n}=440}\]

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Solution: ii

\(\text{given }a = 7, a_{13} = 35,\text{ find }d\text{ and }S_{13}.\) $$\begin{aligned}a_{13}&=a_{1}+\left( 13-1\right) d\\ 35&=7+12d\\ \Rightarrow 12d&=28\\ d&=\dfrac{28}{12}\\ d&=\dfrac{7}{3}\end{aligned}$$ Sum of series $$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ S_{13}&=\dfrac{13}{2}\left[ 2\times 7+\left( 13-1\right) \times \dfrac{7}{3}\right] \\ &=\dfrac{13}{2}\left[ 14+\dfrac{12\times 7}{3}\right] \\ &=\dfrac{13}{2}\left( 14+28\right) \\ &=\dfrac{13}{2}\times 42\\ &=13\times 21\\ &=273\end{aligned}$$ $$\begin{aligned}\boxed{d=\dfrac{7}{3},\ S_{13}=273}\end{aligned}$$

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Solution: iii

Given: \(a_{12}=37,d=3\)
To find \(S_{12}\)

$$\begin{aligned}a_{12}&=a+11d\\ 37&=a+11\times 3\\ a&=37-33\\ a&=4\end{aligned}$$ $$a=4,d=3$$ $$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ S_{12}&=\dfrac{12}{2}\left[ 2\times 4+\left( 12-1\right) 3\right] \\ &=6\left[ 8+11\times 3\right] \\ &=6\left[ 8+33\right] \\ &=6\left( 41\right) \\ &=246\end{aligned}$$

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Solution: iv

\(\text{given } a_3 = 15, S_{10}= 125,\text{ find }d\text{ and }a_{10}.\) $$\begin{align}a_{3}&=15\\ S_{10}&=125\\ a_{3}&=a+2d\\ 15&=a+2d\\ \Rightarrow 2d&=15-a\\ d&=\dfrac{15-a}{2}\tag{1}\end{align}$$ $$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ 125&=\dfrac{10}{2}\left[ 2a+9d\right] \\ 12&5=5\left[ 2a+\dfrac{9\left( 15-a\right) }{2}\right] \\ 25&=\dfrac{4a+135-9a}{2}\\ 50&=135-5a\\ 5a&=135-50\\ 5a&=85\\ a&=\dfrac{85}{5}\\ a&=17\end{aligned}$$ Substituting value of a in equation-(1) $$\begin{aligned}d&=\dfrac{15-\left( 17\right) }{2}\\ &=\dfrac{-2}{2}\\ &=-1\end{aligned}$$ $$\begin{aligned}a_{10}&=a_{1}+9d\\ &=17+9\left( -1\right) \\ &=17-9\\ &=8\end{aligned}$$ \[\boxed{d=-1,\ a_{10}=8}\]

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Solution: v

\(\text{given } d = 5, S_9= 75,\text{ find }a\text{ and }a_9.\)

To Find \(a\) and \(a_9\)

$$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ S_{9}&=\dfrac{9}{2}\left[ 2a+\left( 9-1\right) 5\right] \\ 75\times 2&=9\left( 2a+40\right) \\ 2a+40&=\dfrac{75\times 2}{9}\\ \left( 2a+40\right) &=\dfrac{50}{3}\\ 6a+120&=50\\ 6a&=50-120\\ 6a&=-70\\ a&=-\dfrac{70}{b}\\ a&=\dfrac{-35}{3}\end{aligned}$$ $$\begin{aligned}a_{9}&=a+8d\\ &=\dfrac{-35}{3}+8\times 5\\ &=-\dfrac{35}{3}+40\\ &=\dfrac{-35+120}{3}\\ &=\dfrac{85}{3}\end{aligned}$$ \[\boxed{a=\dfrac{-35}{3},\ \dfrac{85}{3}}\]

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Solution: vi

\(\text{given } a = 2, d = 8, S_n = 90,\text{ find }n\text{ and }a_n .\)

To find \(n\) and \(a_n\)

$$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ 90&=\dfrac{n}{2}\left[ 2\times 2+\left( n-1\right) 8\right] \\ 180&=n\left( 4+8n-8\right) \\ 180&=n\left( 8n-4\right) \\ 180&=8n^{2}-4n\\ \Rightarrow 8n^{2}-4n-180&=0\end{aligned}$$ Dividing Equation by 4 through out $$\scriptsize\begin{aligned}2n^{2}-n-45&=0\\ 2n^{2}-10\eta +9n-45&=0\\ 2n\left( n-5\right) +9\left( n-5\right) &=0\\ \left( n-5\right) \left( 2n+9\right) &=0\\ n-5&=0\\ n&=5\\ 2n+9&=0\\ n &=\frac{-9}{2}\end{aligned}$$ Number of Terms can not be negative, therfore, $$\begin{aligned}n&=5\\ a_{n}&=a+\left( n-1\right) d\\ a_{5}&=2+\left( 4\right) 8\\ &=2+32\\ &=34\end{aligned}$$ $$\boxed{n=5,\; a_{n}=34}$$

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Solution: vii

\(\text{given } a = 8, a_n= 62, S_n= 210,\text{ find }n\text{ and }d.\)

To find \(n\) and \(d\)

$$\begin{align}a_{n}&=a+\left( n-1\right) d\\ 62&=8+\left( n-1\right) d\\ \left( n-1\right) d&=62-8\\ \left( n-1\right) d&=54\\ d&=\dfrac{54}{\left( n-1\right) }\tag{1}\end{align}$$ $$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ 210&=\dfrac{n}{2}\left[ 2\times 8+\left( n-1\right) \cdot \dfrac{54}{\left( n-1\right) }\right] \\ 210&=\dfrac{n}{2}\left[ 16+54\right] \\ 210&=\dfrac{n}{2}\left( 70\right) \\ 210&=35n\\ n&=\dfrac{210}{35}\\ &=6\end{aligned}$$

Substitute the value of \(n\) in equation-(1)

$$\begin{aligned}d&=\dfrac{54}{n-1}\\ &=\dfrac{54}{6-1}\\ &=\dfrac{54}{5}\end{aligned}$$ $$\boxed{n=6,\; d=\dfrac{54}{5}}$$

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Solution: viii

\(\text{given } a_n= 4, d = 2, S_n= –14,\text{ find }n\text{ and }a.\)

To find \(n\) and \(a\)

$$\begin{align}a_{n}&=a+\left( n-1\right) d\\ a_{n}&=a+2\left( n-1\right) \\ 4&=a+2\left( n-1\right) \\ 2\left( n-1\right) &=4-a\\ n-1&=\dfrac{4-a}{2}\\ n&=\dfrac{4-a}{2}+1\\ &=\dfrac{4-a+2}{2}\\ n&=\dfrac{6-a}{2}\tag{1}\end{align}$$ $$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ -14&=\dfrac{6-a}{2\times 2}\left[ 2a+\left( \dfrac{6-a}{2}-1\right) \cdot 2\right] \\ -56&=6-a\left[ 2a+6-a-2\right] \\ -56&=\left( 6-a\right) \left( a+4\right) \\ -5&6=6a+24-a^{2}-4a\\ -56&=2a+24-a^{2}\\ \Rightarrow &a^{2}-2a-24-56=0\\ \Rightarrow &a^{2}-2a-80=0\\ \Rightarrow &a^{2}-10a+8a-80=0\\ &a\left( a-10\right) +8\left( a-10\right) =0\\ &\left( a-10\right) \left( a+8\right) =0\\ &a-10=0\\ &a=10\\ &a+8=0\\ &a=-8\end{aligned}$$

\(a=-8\) or \(a = 10\)

$$n=\dfrac{6-a}{2}$$

when \(a =-8\)

$$\begin{aligned}n&=\dfrac{6-\left( -8\right) }{2}\\ &=\dfrac{14}{2}\\ &=7\end{aligned}$$

when \(a = 10\)

$$\begin{aligned}n&=\dfrac{6-10}{2}\\ &=\dfrac{-4}{2}\\ &=-2\end{aligned}$$

\(n\) can not be negative
therefore,

$$\boxed{n=7,\; a=-8}$$

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Solution: ix

\(\text{given } a = 3, n = 8, S = 192,\text{ find }d.\)

To find \(d\)

$$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ 192&=\dfrac{8}{2}\left[ 2\times 3+7d\right] \\ \dfrac{192}{4}&=6+7d\\ 48&=6+7d\\ 7d&=48-6\\ 7d&=42\\ d&=\dfrac{42}{7}\\ &=6\\ d&=6\end{aligned}$$ \[\boxed{d=6}\]

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Solution: x

\(\text{given } l = 28, S = 144,\text{ and there are total 9 terms. Find } a.\) $$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ a+l\right] \\ 144&=\dfrac{9}{2}\left[ a+28\right] \\ 288&=9a+252\\ 9a&=288-252\\ 9a&=36\\ a&=\dfrac{36}{9}\\ &=4\end{aligned}$$ \[\boxed{a=4}\]

Q4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Solution

AP: 9, 17, 25, ... $$\begin{aligned}a&=9\\ d&=17-9\\ &=8\end{aligned}$$

Let \(n\)th term of AP sum up to 636

$$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ 636&=\dfrac{n}{2}\left[ 2\times 9+\left( n-1\right) 8\right] \\ &=\dfrac{n}{2}\left( 18+8n-8\right) \\ &=\dfrac{n}{2}\left( 10+8n\right) \\ 636&=5n+4n^{2}\\ \Rightarrow &4n^{2}+5n-636=0\\ \Rightarrow &4n^{2}+53n-48n-636=0\\ \Rightarrow &n\left( 4n+53\right) -12\left( 4n+53\right) =0\\ &\left( 4n+53\right) \left( n-12\right) =0\\ n-12&=0\\ n&=12\\ 4n+53&=0\\ n&=-53/4\end{aligned}$$

No of terms in AP cannot be less than zero or a non-integer
therefore,

$$\boxed{n=12}$$ 12 terms should be taken to obtain required sum =636

Q5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution:

First term of AP \(a=5\)
last term of AP \(l=45\)
Sum of terms \(S=400\)

$$\begin{aligned}S=400\\ S&=\dfrac{n}{2}\left[ a+l\right] \\ 400&=\dfrac{n}{2}\left[ 5+45\right] \\ 400&=\dfrac{50n}{2}\\ \Rightarrow n&=\dfrac{400}{25}\\ n&=16\\ a_{n}&=a_{1}+\left( n-1\right) d\\ 45&=5+15d\\ \Rightarrow 15d=40\\ d&=\dfrac{40}{15}\\ &=\dfrac{8}{3}\end{aligned}$$ $$\boxed{n=16,\; d=\dfrac{8}{3}}$$

Q6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution:

First term of AP \(a=17\)
last term of AP \(l=350\)
Common difference \(d=9\)

$$\begin{aligned}a_{n}&=a+\left( n-1\right) d\\ 350&=17+\left( n-1\right) 9\\ 350&=17+9n-9\\ 350&=8+9n\\ \Rightarrow 9n&=350-8\\ 9n&=342\\ n=\dfrac{342}{9}\\ n&=38\end{aligned}$$ Sum of series $$\begin{aligned}S&=\dfrac{n}{2}\left[ a+l\right] \\ &=\dfrac{38}{2}\left[ 17+350\right] \\ &=19\times 367\\ &=6973\end{aligned}$$ $$\boxed{n=38,\; S=6973}$$

7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Solution:

Common Difference \(d=7\)
and 22nd term \(a_{22}=149\)

$$\begin{aligned}a_{22}&=a+\left( n-1\right) d\\ 149&=a+\left( 22-1\right) 7\\ 149&=a+21\times 7\\ 149&=a+147\\ a&=149-147\\ &=2\end{aligned}$$ Sum of first 22 terms of AP $$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ S_{22}&=\dfrac{22}{2}\left[ 2\cdot 2+21\times 7\right] \\ &=11\left[ 4+147\right] \\ &=11\times 151\\ &=1661\end{aligned}$$ Sum of the first 22 terms of the AP \[\text{1661}\]

Q8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution:

Second and third term of AP is 14 and 18 respectively $$\begin{aligned}d&=a_{3}-a_{2}\\ d&=18-14\\ &=4\end{aligned}$$ Sum of first 51 terms $$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ S_{51}&=\dfrac{51}{2}\left[ 2\times 10+\left( 51-1\right) \times 4\right] \\ &=\dfrac{51}{2}\left( 20+50\times 4\right) \\ &=\dfrac{51}{2}\left( 20+200\right) \\ &=\dfrac{51}{2}\times 220\\ &=51\times 110\\ &=5610\end{aligned}$$ The sum of first 51 terms of an AP \[\boxed{5610}\]

Q9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first \(n\) terms.

Solutions:

Sum of first 7 terms = 49
and sum of first 17 term = 289
to find sum of \(n\) terms

$$\scriptsize\begin{align}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ S_{7}&=\dfrac{7}{2}\left[ 2a+6d\right] \\ 49&=7\left( a+3d\right) \\ 7&=a+3d\tag{1}\end{align}$$ $$\scriptsize\begin{align}S_{17}&=\dfrac{17}{2}\left[ 2a+16d\right] \\ 289&=17\left( a+8d\right) \\ 17&=a+8d\tag{2}\end{align}$$ Subtracting equation-(1) from equation-(2) $$ \begin{aligned} a+8d-a-3d&=17-7\\ 5d&=10\\ d&=\dfrac{10}{5}\\ &=2 \end{aligned} $$

Substituting value of \(d =2\) in equation-(1)

$$\scriptsize\begin{aligned}7&=a+3d\\ 7&=a+3\times 2\\ a&=7-6\\ &=1\\\\ S_{n}&=\dfrac{n}{2}\left[ 2\times 1+\left( n-1\right) 2\right] \\ &=\dfrac{n}{2}\left[ 2+2n-2\right] \\ S_{n}&=n^{2}\end{aligned}$$

Sum of \(n\) terms=\[\boxed{S_{n}=n^{2}}\]

Q10. Show that \(a_1,\ a_2,\ . . .,\ a_n, . . .\) form an AP where an is defined as below :
(i) \(a_n= 3 + 4n\)
(ii) \(a_n= 9 – 5n\)
Also find the sum of the first 15 terms in each case.

Solution: (i) \(a_n= 3 + 4n\)

$$\begin{aligned}a_{n}&=3+4n\\ a_{1}&=3+4=7\\ a_{2}&=3+4\times 2=11\\ a_{3}&=3\times 4\times 3=15\\ 7,11,15....\\ d&=a_{3}-a_{2}=a_{2}-a_{1}\\ &=15-11=11-7\\ &=4=4\end{aligned}$$

Hence, \(a_{n}=3+4n\) forms an AP

Sum of 15 term $$\scriptsize\begin{aligned}S_{15}&=\dfrac{15}{2}\left[ 2\cdot 7+\left( 15-1\right) 4\right] \\ &=\dfrac{15}{2}\left[ 14+14\times 4\right] \\ &=\dfrac{15}{2}\left[ 14+56\right] \\ &=\dfrac{15}{2}\times 70\\ &=15\times 35\\ &=525\end{aligned}$$ Sum of 15 terms \[\boxed{S_{15}=525}\]

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Solution: (ii) \(a_n= 9 – 5n\)

$$\begin{aligned}a_{n}&=9-5n \\ a_{1}&=9-5=4\\ a_{2}&=9-10=-1\\ a_{3}&=9-15=-6\\ a_{4}&=9-20=-11\end{aligned}$$ $$\text{AP: }4,-1,-6,-11$$ $$\scriptsize\begin{aligned}d=a_{4}-a_{3}=-11-\left( -6\right)=-5&\\ =a_{3}-a_{2}=-6-\left( -1\right) =-5\\ =a_{2}-a_{1} =-1-\left( 4\right)=-5\\ \end{aligned}$$

Common Difference \(d=5\) is same across terms, therefore \(a_n= 9 – 5n\) is in AP

$$\scriptsize\begin{aligned}S_{15}&=\dfrac{15}{2}\left[ 2\times 4+\left( 15-1\right) \left( -5\right) \right] \\ &=\dfrac{15}{2}\left[ 8+14\times \left( -5\right) \right] \\ &=\dfrac{15}{2}\left( 8-70\right) \\ &=-\dfrac{15\times 62}{2}\\ &=-15\times 31\\ &=-465\end{aligned}$$ \[\boxed{S_{15}=-465}\]

Q11. If the sum of the first \(n\) terms of an AP is \(4n – n^2\), what is the first term (that is \(S_1\) )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Solution:

$$\begin{aligned}S_{n}&=4n-n^{2}\\ S_{1}&=4\cdot 1-1^{2}\\ &=4-1\\ &=3\\\\ S_{2}&=4\cdot 2-2^{2}\\ &=8-4\\ &=4\end{aligned}$$ $$\begin{aligned}a_{2}&=S_{2}-S_{1}\\ &=4-3\\ &=1\\\\ a_{3}&=S_{3}-S_{2}\\\\ S_{3}&=4\times 3-3^{2}\\ &=12-9\\ &=3\\\\ a_{3}&=3-4\\ &=-1\end{aligned}$$ $$\begin{aligned}a_{10}&=S_{10}-S_{9}\\ S_{10}&=4\times 10-10^{2}\\ &=4-100\\ &=-60\\\\ S_{9}&=4\times 9-9^{2}\\ &=36-81\\ &=-45\\\\ a_{10}&=-60-\left( -45\right) \\ &=-15\end{aligned}$$ $$\begin{aligned} S_{n}&=4n-n^{2}\\\\ S_{n-1}&=4\left( n-1\right) -\left( n-1\right) ^{2}\\ &=4n-4-\left( n^{2}+1-2n\right) \\ &=4n-4-n^{2}-1+2n\\ &=6n-5-n^{2}\\\\ a_{n}&=S_{n}-S_{n-1}\\ &=4n-n^{2}-6n+5+n^{2}\\ &=5-2n\end{aligned}$$

Q12. Find the sum of the first 40 positive integers divisible by 6.

Solution:

First 40 positive nos divisible by 6 are in form of AP
6, 12, 18 .... to 40 terms $$\scriptsize\begin{aligned}a&=6\\ d&=12-6=6\\ n&=40\\\\ S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\\\ S_{40}&=\dfrac{40}{2}\left[ 2\times 6+\left( 40-1\right) 6\right] \\ &=20\left[ 12+39\times 6\right] \\ &=20\left( 12+234\right) \\ &=20\times 246\\ &=4920\end{aligned}$$

Sum of the first 40 positive integers divisible by 6 \[\boxed{S_{40}=4920}\]

Q13. Find the sum of the first 15 multiples of 8.

Solution:

Multiple of 8 are in AP as $$8,\ 16,\ 24,\ 32,\ \ldots$$

First term \(a=8\)
Common Difference \(d=8\)

$$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\\\ S_{15}&=\dfrac{15}{2}\left[ 2\times 8+\left( 15-1\right) 8\right] \\ &=\dfrac{15}{2}\left[ 16+14\times 8\right] \\ &=\dfrac{15}{2}\left( 16+112\right) \\ &=\dfrac{15}{2}\times 128\\ &=15\times 64\\ &=960\end{aligned}$$

Sum of the first 15 multiples of 8 \[\boxed{S_{15}=960}\]

Q14. Find the sum of the odd numbers between 0 and 50.

Solution:

AP Formed by odd numbers is as $$\begin{aligned}1,\ 3,\ 5,\ 7,\ \ldots ,\ 49\end{aligned}$$

First term \(a=1\)
Common Difference \(d=2\)
Last term \(a_n=49\)

$$\begin{aligned}a_{n}&=a+\left( n-1\right) d\\ 49&=1+\left( n-1\right) \times 2\\ 49&=1+2n-2\\ 2n&=50\\ n&=25\end{aligned}$$ Sum of odd numbers between 0-50 $$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\ S_{25}&=\dfrac{25}{2}\left[ 2\times 1+\left( 25-1\right) \cdot 2\right] \\ &=\dfrac{25}{2}\left( 2+48\right) \\ &=\dfrac{25\times 50}{2}\\ &=625\end{aligned}$$

Q15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ` 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Solution:

AP Formed by Penalties $$\begin{aligned}200,250,300,...\end{aligned}$$

First term \(a=200\)
Common Difference \(d=50\)
Number of terms \(n=30\)

$$\scriptsize\begin{aligned}S_{n}&=\dfrac{n}{2}\left[ 2a+\left( n-1\right) d\right] \\\\ S_{30}&=\dfrac{30}{2}\left[ 2\times 200+\left( 29\right) 50\right] \\ &=15\left[ 400+29\times 50\right] \\ &=15\left( 400+1450\right) \\ &=15\times 1850\\ &=27750\end{aligned}$$

Penalty amount that contractor paid \[\boxed{\mathbb{\large{\text{₹ }}} 27,750.00}\]

Q16. A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.

Solution:

Let Highest Cash prize be\(x\)

Series of prizes will be $$x,x-20,x-40,....$$ $$\begin{aligned}a&=x\\ d&=-20\\ n &=7\\\\ a_{n}&=a+\left( n-1\right) d\\ a_{7}&=x+6\times \left( -20\right) \\ &=x-120\end{aligned}$$ Sum of series of Prizes $$\scriptsize\begin{aligned}S&=\dfrac{\eta }{2}\left( a+l\right) \\ 700&=\dfrac{7}{2}\left[ x+x-120\right] \\ 200&=2x-120\\ 2x&=200+120\\ 2x&=320\\ x&=160\end{aligned}$$ Prizes will be $$\text{₹}160,\; \text{₹}140,\; \text{₹}120,\; \text{₹}100,\; \text{₹}80,\; \text{₹}60,\; \text{₹}40$$

Q17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution:

AP Formed by plantation of tree $$1,2,3,....12$$

First term \(a=1\)
Common Differenec \(d=1\)
Last term \(l=12\)

Sum of plants $$\begin{aligned}S&=\dfrac{12}{2}\left[ a+l\right] \\ &=6\left( 1+12\right) \\ &=6\times 13\\ &=78\end{aligned}$$

But there are 3 sections of one class, therfore, No of plants would be 3 times

$$\begin{aligned}&=78\times 3\\ &=234\end{aligned}$$

No of plants\[\boxed{234}\]