ARITHMETIC PROGRESSIONS-Exercise 5.2

Q1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP: \[ \begin{array}{|c|c|c|c|c|} \hline &a&d&n&a_n\\\hline i.&7&3&8&\color{red}\underline{28}\\\hline ii.&-18&\color{red}\underline{2}&10&0\\\hline iii.&\color{red}\underline{46}&-3&18&-5\\\hline iv.&-18.9&2.5&\color{red}\underline{10}&3.6\\\hline v.&3.5&0&105&\color{red}\underline{3.5}\\\hline \end{array} \]

Solution:(i)

$$\begin{aligned} a&=7\\ d&=3\\ n&=8\\ a_{n}&=a+\left( n-1\right) d\\ &=7+\left( 8-1\right) 3\\ &=7+7\times 3\\ &=28\end{aligned}$$

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Solution:(ii)

$$\begin{aligned} a&=-18\\ n&=10\\ a_{n}&=0\\ a_{n}&=a+\left( n-1\right) d\\ 0&=-18+\left( 10-1\right) d\\ \Rightarrow 9d&=18\\ d&=\dfrac{18}{9}\\ &=2\end{aligned}$$

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Solution:(iii)

$$\begin{aligned}d=-3\\ n&=18\\ 9n&=-5\\ a_{n}&=a+\left( n-1\right) d\\ -5&=a+\left( 18-1\right) \left( -3\right) \\ \Rightarrow a&=-5+51\\ a&=46\end{aligned}$$

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Solution:(iv)

$$\begin{aligned} a&=-18.9\\ d&=2.5\\ a_{n}&=3.6\\ a_{n}&=a+\left( n-1\right) d\\ 3.6&=-18.9+\left( n-1\right) 2.5\\ \Rightarrow \left( n-1\right) 2\cdot 5&=3.6+18.9\\ &=22.5\\ n-1&=\dfrac{22.5}{2.5}\\ &=9\\ n&=9+1\\ &=10\end{aligned}$$

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Solution:(v)

$$\begin{aligned} a&=3.5\\ d&=0\\ n&=105\\ a_{n}&=a_{1}+\left( n-1\right) d\\ &=3.5T\left( 105-1\right) \times 0\\ &=3.5+104\times 0\\ &=3.5\end{aligned}$$

Q2.Choose the correct choice in the following and justify :

1. 30th term of the AP: 10, 7, 4, . . . , is
   \( (A)\ 97\qquad (B)\ 77 \qquad (C)\ –77\ \color{red}{\checkmark} \color{white}\qquad (D)\ – 87\)
2. 11th term of the AP: \(-3,\ -\frac{1}{2},\ 2,\ \ldots\)
   \( (A)\ 28\qquad (B)\ 22 \color{red}{\checkmark} \color{white} \qquad (C)\ -38\qquad (D)\ – 48\frac{1}{2}\)

Solution:(i)

\(\text{AP: }10,\ 7,\ 4,\ \ldots\) $$\begin{aligned} a_1&=10\\ a_2&=7\\ d&=a_{2}-a_{1}\\ &=7-10\\ &=-3\\\\ a_{30}&=a+\left( n-1\right) d\\ &=10+\left( 30-1\right) \left( -3\right) \\ &=10+29\times \left( -3\right) \\ &=10-87\\ &=-77\end{aligned}$$

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Solution:(i)

\(\text{AP: }-3,\ -\dfrac{1}{2},\ 2,\ \ldots \) $$\begin{aligned} a_{1}&=-3,\\ a_2&=-\dfrac{1}{2}\\ d&=a_{2}-a_{1}\\ &=-\dfrac{1}{2}-\left( -3\right) \\ &=-\dfrac{1}{2}+3\\ &=\dfrac{5}{2}\\\\ a_{11}&=a_{1}+\left( n-1\right) d\\ &=-3+\left( 11-1\right) \dfrac{5}{2}\\ &=-3+10\times \dfrac{5}{2}\\ &=-3+25\\ a_{11}&=22\end{aligned}$$

Q3. In the following APs, find the missing terms in the boxes : \[ \begin{array}{} i.&2,&\boxed{\color{red}{14}},& 26\\ ii.&\boxed{\color{red}{18}},&13,&\boxed{\color{red}{8}},&3\\ iii.&5,&\boxed{\color{red}{6\frac{1}{2}}},&\boxed{\color{red}{8}},&9\frac{1}{2}\\ iv&-4,&\boxed{\color{red}-2},&\boxed{\color{red}0},&\boxed{\color{red}2},&\boxed{\color{red}4},&6\\ v.&\boxed{\color{red}53},&38,&\boxed{\color{red}23},&\boxed{\color{red}8},&\boxed{\color{red}-7},&-22 \end{array} \]

Solution (i):

\(2,\ \ldots,\ 26\) $$\begin{aligned} a&=2\\ a_{3}&=26\\ a_{3}&=a_{1}+\left( 3-1\right) d\\ 26&=2+2d\\ 2d&=24\\ d&=\frac{24}{2}\\ &=12\\\\ a_{2}&=a_{1}+d\\ &=2+12\\ &=14\end{aligned}$$

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Solution (ii):

\(\ldots,\ 13,\ \ldots,\ 3\) $$\begin{aligned} a_{2}&=13\\ \Rightarrow a_{1}+d&=13\\ a_{1}+3d&=3\\ -2d&=10\\ d&=-5\\\\ a_{2}&=a_{1}+d\\ 13&=a_{1}-5\\ a_{1}&=13+5\\ &=18\\\\ a_{3}&=a_{1}+2d\\ &=18+2\left( -5\right) \\ &=18-10\\ &=8\end{aligned}$$

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Solution (iii):

\(5,\ -1\ \ldots,\ 9\dfrac{1}{2}\) $$\begin{aligned} a_1&=5\\ a_{4}&=a_{1}+3d\\ \dfrac{19}{2}&=5+3d\\ 3d&=\dfrac{19}{2}-5\\ &=\dfrac{19-10}{2}\\ &=\dfrac{9}{2}\\ d&=\dfrac{3}{2}\\\\ a_{2}&=a_{1}+d\\ &=5+\dfrac{3}{2}\\ &=\dfrac{13}{2}\\ &=6\dfrac{1}{2}\\\\ a_{3}&=a_{1}+2a\\ &=5+2\cdot \dfrac{3}{2}\\ &=8\end{aligned}$$

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Solution (iv):

\(-4,\ \ldots,\ \ldots,\ \ldots,\ \ldots,\ 6\) $$\begin{aligned} a_{1}&=-4\\ a_{6}&=6\\ a_{6}&=a_{1}+5d\\ 6&=-4+5d\\ 5d&=10\\ d&=\dfrac{10}{5}\\ d&=2\end{aligned}$$ $$\begin{aligned} a_{1}&=-4\\ d&=2\\ a_{2}&=a_{1}+d\\ &=-4+2\\ &=-2\\\\ a_{3}&=a_{1}+2d\\ &=-4+\left( 2\times 2\right) \\ &=0\\\\ a_{4}&=a_{1}+3d\\ &=-4+3\times 2\\ &=2\\\\ a_{5}&=a_{1}+4d\\ &=-4+4\times 2\\ &=4\end{aligned}$$

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Solution (iv):

\(\ldots,\ 38,\ \ldots,\ \ldots,\ \ldots,\ -22\) $$\begin{align} a_{2}&=38\\ a_{6}&=-22\\ a_{2}&=a_{1}+d\\ 38&=a_{1}+d\tag{1}\end{align}$$ $$\begin{align} a_{6}&=a_{1}+5d\\ -22&=a_{1}+5d\tag{2}\end{align}$$ Subtracting equation-(1) from equation(2) $$\begin{aligned} a_{1}+5a-\left( a_{1}+d\right) &=-22-38\\ 4d&=-60\\ d&=-15\\\\ a_{1}&=a_{2}-d\\ &=38-\left( -15\right) \\ &=38+15\\ &=53\\\\ a_{3}&=a_{1}+2d\\ &=53+2\left( -15\right) \\ &=53-30\\ &=23\\\\ a_{4}&=a_{1}+3d\\ &=53+3\left( -15\right) \\ &=53-45\\ &=8\\\\ a_{5}&=a_{1}+4d\\ &=53+4\left( -15\right) \\ &=53-60\\ &=-7\end{aligned}$$

Q4. Which term of the AP : 3, 8, 13, 18, . . . ,is 78?

Solution:

\(\text{AP: }3,\ 8,\ 13,\ 18,\ \ldots,\ 78\) $$\begin{aligned} a_{1}&=3\\ d&=92-a_{1}\\ &=8-3\\ &=5\\\\ a_{n}&=78\\ a_{n}&=a_{1}+\left( n-1\right) d\\ 78&=3+\left( n-1\right) 5\\ \left( n-1\right) 5&=78-3\\ &=75\\ \left( n -1\right) &=\dfrac{75}{5}\\ &=15\\ n-1&=15\\ n &=15+1\\ &=16\end{aligned}$$ 78 is 16th term.

Q5. Find the number of terms in each of the following APs :

• 7, 13, 19, . . . , 205
• 18, \(15\frac{1}{2}, 13, ....., -47\)

Solution (i):

\(7,\ 13,\ 19,\ \ldots,\ 205\) $$\begin{aligned} a_{1}&=7\\ d&=a_{2}-a_{1}\\ &=13-7\\ &=6\\\\ a_{n}&=a_{1}+\left( n-1\right) d\\ 205&=7+\left( n-1\right) 6\\ 6\left( n-1\right) &=205-7\\ &=198\\ \left( n-1\right) &=\dfrac{198}{6}\\ \left( n-1\right) &=33\\ n&=33+1\\ &=34\end{aligned}$$ 205 is 34th term in the AP

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Solution (ii):

\(18,\ 15\dfrac{1}{2},\ 13,\ \ldots,\ -47\) $$\begin{aligned} a_{1}&=18\\ a_{2}&=15\dfrac{1}{2}\\ d&=a_{2}-a_{1}\\ &=15\dfrac{1}{2}-18\\ &=\dfrac{31}{2}-18\\ &=\dfrac{31-36}{2}\\ &=\dfrac{-5}{2}\\\\ a_{n}&=-47\\ a_{n}&=a_{1}+\left( n-1\right) d\\ 18+\left( n-1\right) d&=-47\\ -\dfrac{5}{2}\left( n-1\right) &=-47-18\\ &=-65\\ \left( n-1\right) &=\dfrac{-65\times 2}{-5}\\ \left( n-1\right) &=26\\ n&=26+1\\ &=27\end{aligned}$$ - 47 is the 27th term in AP

Q6. Check whether –150 is a term of the AP : 11, 8, 5, 2 . . .

Solution:

\(\text{AP: }11,\ 8,\ 5,\ 2,\ ....,\ -150\) $$\text{Let }a_{n}=-150$$ $$\begin{aligned} a_{1}&=11\\ a_{2}&=8\\ d&=92-a_{1}\\ &=8-11\\ &=-3\\\\ a_{n}&=a_{1}+\left( n-1\right)\\ a_{1}+\left( n-1\right) \times \left( -3\right) &=-150\\ 11+\left( n-1\right) \times 3&=150\\ 3\left( n-1\right) &=150-11\\ 3\left( n-1\right) &=139\\\\ \left( n-1\right) &=\dfrac{139}{3}\\\\ n&=\dfrac{139-3}{3}\\\\ n&=\dfrac{136}{3}\\\\ \Rightarrow n &\notin \mathbb{Z}\end{aligned}$$ -150 is not in AP as value of n is not an integer

Q7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution:

$$\begin{align} a_{11}&=38\\ a_{16}&=73\\ a_{11}&=a_{1}+\left( 11-1\right) d=38\\ &=a_{1}+10d=38\tag{1}\end{align}$$ $$\begin{align} a_{16}&=a_{1}+\left( 16-1\right) d\\ &=a_{1}+15d=73\tag{2}\end{align}$$ Subtracting Equation-(1) from Equation-(2) $$\begin{aligned} a_{1}+15d-\left( a_{1}+10d\right) &=73-38\\ 5d&=35\\ d&=\dfrac{35}{5}\\ &=7\end{aligned}$$

Substituting \(d = 7\) in to esuation-(1)

$$\begin{aligned} a_{1}+10d&=38\\ a_{1}+10\times 7&=38\\ a_{1}&=38-70\\ &=-32\\\\ a_{31}&=a_{1}+\left( 31-1\right) \times 7\\ &=-32+30\times 7\\ &=-32+210\\ &=178\end{aligned}$$ 31th term in the AP is 178

Q8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

$$\begin{align} a_{3}&=12\\ a_{50}&=106\\ a_{3}&=a_{1}+\left( n-1\right) d\\ &=a_{1}+\left( 3-1\right) d\\ &=a_{1}+2d=12\tag{1}\end{align}$$ $$\begin{align} a_{50}&=a_{1}+\left( n-1\right) d=106\\ &=a_{1}+\left( 50-1\right) d=106\\ &=a_{1}+49d=106\tag{2}\end{align}$$ Subtracting equation-(1) from equation-(2) $$\begin{aligned} \left( a_{1}+49d\right) -\left( a_{1}+2d\right) &=106-12\\ 47d&=94\\ d&=\dfrac{94}{47}\\ &=2\end{aligned}$$

Substitute \(d=2\) into equation-(1)

$$\begin{aligned} a_{1}+2d&=12\\ a_{1}+2\cdot 2&=12\\ a_{1}&=12-4\\ a_{1}&=8\\\\ a_{1}=8,&\ d=2\\\\ a_{29}&=a_{1}+\left( 29-1\right) \times 2\\ &=8+28\times 2\\ &=8+56\\ &=64\end{aligned}$$ 29th term is 64

Q9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

Solution:

$$\begin{align} a_{3}&=4\\ a_{9}&=-8\\ a_{3}&=a_{1}+\left( 3-1\right) d\\ \Rightarrow a_{1}+2d&=4\tag{1}\end{align}$$ $$\begin{align} a_{9}&=a_{1}+\left( 9-1\right) d\\ \Rightarrow a_{1}+8d&=-8\tag{2}\end{align}$$ Subtracting equation-(1) from equation-(2) $$\begin{aligned} \left( a_{1}+8d\right) -\left( a_{1}+2d\right) &=-8-4\\ 8d-2d&=-12\\ 6d&=-12\\ d&=\dfrac{-12}{6}\\ &=-2\end{aligned}$$

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

$$\begin{aligned}a_{1}+2d&=4\\ a_{1}+2\left( -2\right) &=4\\ a_{1}&=8\\\\ a_{1}=8,\ &d=-2\end{aligned}$$ let nth term is 0 $$\begin{aligned} a_{n}&=a_{1}+\left( n-1\right) d\\ \Rightarrow 8+\left( n-1\right) \left( -2\right) &=0\\ \left( n-1\right) \left( -2\right) &=-8\\ n-1&=4\\ n&=5\end{aligned}$$ 5th term in AP is zero

Q10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution:

$$\begin{aligned} a_{17}&=a_{1}+16d\\ a_{10}&=a_{1}+9d\end{aligned}$$

17th term \((a_{17})\) exceeds tenth term (\(a_{10}\)) by 7, therefore

$$\begin{aligned} a_{1}+9d+7&=a_{1}+16d\\ 9d+7&=16d\\ 16d-9d&=7\\ 7d&=7\\ d&=1\end{aligned}$$

Common Difference \(d=1\)

Q11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Solution:

\(\text{AP: }3,\ 15,\ 27,\ 39,\ \ldots \) $$\begin{aligned} a_{1}&=3\\ a_{2}&=15\\ d&=a_{2}-a_{1}\\ &=15-3\\ &=12\\\\ a_{54}&=a_{1}+\left( 53-1\right) d\\ &=3+52\times 12\\ &=3+624\\ &=627\end{aligned}$$

Let nth term be 132 more than 54th term

$$\begin{aligned} a_{n}&=a_{1}+\left( n-1\right) 12\\ 627+132&=3+\left( n-1\right) 12\\ 759-3&=\left( n-1\right) 12\\ 12\left( n-1\right) &=756\\ \left( n-1\right) &=\dfrac{756}{12}\\ \left( n-1\right) &=63\\ n&=64\end{aligned}$$ 64th term in the AP is 132 more than 54th term

Q12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution:

Let first term of first AP $$a_{f}$$ and first term of Second AP $$a_{s}$$ Common difference $$=d$$ 100th term of each AP $$\begin{aligned}t_{1}&=a_{f}+99d\\ t_{2}&=a_{s}+99d\end{aligned}$$ Difference between 100th term $$\begin{aligned} a_{f}+99d-\left( a_{s}+99d\right) &=100\\ a_{f}-a_{s}&=100\end{aligned}$$ 1000th term of both AP $$\begin{aligned} S_{1}&=a_{f}+999d\\ S_{2}&=a_{s}+999d\\ S_{1}-S_{2}&=a_{t}-a_{s}\\ &=100\end{aligned}$$ therefore, difference remain same which is 100

Q13. How many three-digit numbers are divisible by 7?

Solution:

3 digits number divisible by 7 are represented in in an AP as $$105,\ 112,\ ...,\ 994$$

Common difference \(d=7\)
first 3 digit divisible by 7

$$a_{1}=105$$ last 3 digit divisible by 7 $$a_{n}=994$$ $$\begin{aligned} a_{n}&=a_{1}+\left( n-1\right) d\\ 994&=105+\left( n-1\right) 7\\ \left( n-1\right) 7&=994-105\\ 7\left( n-1\right) &=889\\ \left( n-1\right) &=\dfrac{889}{7}\\ &=127\\ \left( n-1\right) &=127\\ n&=128\end{aligned}$$ there are 128 three digits number divisible by 7

Q14. How many multiples of 4 lie between 10 and 250?

Solution:

Multiples of 4 between 10 and 250 are arrnged in an AP as $$12,\ 16,\ ....,\ 248$$ $$\begin{aligned} a_{1}&=12\\ d&=4\\ a_{n}&=248\\\\ a_{n}&=a_{1}+\left( n-1\right) d\\ 248&=12+\left( n-1\right) 4\\ 4\left( n-1\right) &=248-12\\ 4\left( n-1\right) &=236\\ n-1&=\dfrac{236}{4}\\ n-1&=59\\ n&=59+1\\ &=60\end{aligned}$$ There are 60 number between 10 to 250 divisible by 4

Q15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Solution:

Consider first AP: $$63,\ 65,\ 67,\ ....$$ $$\begin{aligned} a_{1}&=63\\ a_2&=65\\ d&=a_{2}-a_{1}\\ &=65-63\\ &=2 \end{aligned}$$ and second AP $$3,\ 10,\ 17,\ \ldots$$ $$\begin{aligned} d'&=3\\ a'_1&=3\\ a'_2&=10\\ d'&=a'_{2}-a_{1}'\\ &=10-3\\ &=7\end{aligned}$$

Let nth term of both APs are equal
nth term of first AP

$$a_{n}=63+\left( n-1\right) 2$$ and nth term of second AP $$a'n=3+\left( n-1\right) 7$$ $$a_{n}=a'_{n}$$ $$ \begin{aligned} 63+\left( n-1\right) 2&=3+\left( n-1\right) 7\\ 63-3&=\left( n-1\right) 7-\left( n-1\right) 2\\ 60&=7n-7-2n+2\\ 60&=5n-5\\ \Rightarrow 5n&=60+5\\ n&=\dfrac{65}{5}\\ n&=13 \end{aligned} $$ 13th term of both APs are equal

Q16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution:

$$\begin{align} a_{3}&=16\\ a_{3}&=a_{1}+2d\\ \Rightarrow a_{1}+2d&=16\tag{1}\end{align}$$

7th term of AP \(a_7\) exceeds the 5th term \(a_5\) by 12

$$\begin{aligned} a_{7}&=a_{1}+6d\\ a_{5}&=a_{1}+4d\\\\ a_{7}&=a_{5}+12\\ \Rightarrow a_{1}+6d&=a_{1}+4d+12\\ \Rightarrow 2d&=12\\ d&=\dfrac{12}{2}\\ &=6\end{aligned}$$

Substituting value of \(d=6\) in equation-(1)

$$\begin{aligned} a_{1}+2d&=16\\ a_{1}&=16-2d\\ &=16-2\times 6\\ a_{1}&=4\end{aligned}$$ AP $$\begin{aligned} &(a_{1}),\ (a_{1}+d),\ (a_{1}+2d),\ (a_{1}+3d),\ \ldots \\ &(4),\ (4+6),\ (4+2\times 6),\ (4+3\times 6),\ \ldots \\ &4,\ 10,\ 16,\ 22,\ \ldots\end{aligned}$$

Q17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

Solution:

$$3,\ 8,\ 13,\ \ldots ,\ 253$$ $$ \begin{aligned} a_{1}&=3\\ a_{2}&=8$\\ d&=a_{2}-a_{1}\\ &=8-3\\ &=5 \end{aligned} $$ $$ \begin{aligned} a_{n}&=a_{1}+\left( n-1\right) d\\ 253&=3+\left( n-1\right) 5\\ 250&=\left( n-1\right) 5\\ \Rightarrow n-1&=\dfrac{250}{5}\\ n-1&=50\\ n =50+1\\ n=51 \end{aligned} $$

20th term from last term would be 32nd froms start \(\left( 51-20\right) +1=32\)

$$\begin{aligned} a_{32}&=a_{1}+\left( n-1\right) d\\ a_{32}&=3+\left( 32-1\right) \times 5\\ &=3+31\times 5\\ &=3+155\\ a_{32}&=158 \end{aligned} $$

20th term from last term is 158

2nd Method

Consider last term as first term thus AP becomes reverse of original with common diffrence with negative sign \(d=-5\)

$$a_{1}=253$$ $$d=-5$$ $$ \begin{aligned} a_{20}&=a_{1}+\left( n-1\right) d\\ &=253+\left( 19\right) \left( -5\right)\\ &=253-195\\ &=158 \end{aligned} $$

Q18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution:

Sum of 4th and 8th term of an AP \(=24\)

$$\begin{align} a_{4}&=a_{1}+3d\\ a_{8}&=a_{1}+7d\\ a_{4}+a_{8}&=24\\ \Rightarrow a_{1}+3d+a_{1}+7d&=24\\ 2a_{1}+10d&=24\tag{1}\end{align}$$

Sum of 6th and 10th term of AP is \(44\)

$$\begin{align} a_{6}&=a_{1}+5d\\ a_{10}&=a_{1}+9d\\ a_{6}+a_{10}&=44\\ \Rightarrow a_{1}+5d+a_{1}+9d&=44\\ \Rightarrow 2a_{1}+14d&=44\tag{2}\end{align}$$

Subtract equation-(1) from equation-(2)

$$\begin{aligned} 2a_{1}+14d-\left( 2a_{1}+10d\right) &=44-24\\ 14d-10d&=20\\ 4d&=20\\ d&=\dfrac{20}{4}\\ &=5\end{aligned}$$

Substituting value of \(d=5\) in equation-(1)

$$\begin{aligned} 2a_{1}+10d&=24\\ 2a_{1}+10\times 5&=24\\ 2a_{1}&=24-50\\ 2a_{1}&=-26\\ a_{1}&=-\dfrac{26}{2}\\ a_{1}&=-13\\ a_{1}&=-13,d=5\end{aligned}$$

APs First three terms

$$\begin{aligned} a_{1}&=-13\\\\ a_{2}&=a_{1}+d\\ &=-13+5\\ &=-8\\\\ a_{3}&=a_{1}+2d\\ &=-13+10\\ &=-3\\\\ -13,\ -8,\ -3\end{aligned}$$

Q19. Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

Solution:

Annual Salary of Subha Rao in 1995 = ₹5,000
Annual increment = ₹200
to find when did he reach to income of ₹7000,
Consider first term (\(a_1\)) as ₹5,000
Common difference (\(d\)) as ₹200
and \(n\)th term (\(a_n\)) = ₹7,000

$$\begin{aligned} a_{1}&=5000\\ d&=200\\ a_{n}&=7000\end{aligned}$$

find \(n\)

$$\begin{aligned} a_{n}&=a_{1}+\left( n-1\right) d\\ 7000&=5000+\left( n-1\right) 200\\ 200\left( n-1\right) &=7000-5000\\ 200\left( n-1\right) &=2000\\ \left( n-1\right) &=\dfrac{2000}{200}\\ \left( n-1\right) &=10\\ n&=11\end{aligned}$$ Subha Rao will receive ₹7,000 in 11th year

Q20. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.

Solution:

Ram Kali first weekly saving = ₹ 5 She increased weekly saving by ₹ 1.75 Her saving grew to ₹ 20. 75 in the nth week

Let First term \(a_1 =5\)
and last term (\a_n=20.75\)
Common Diffrenece \(d=1.75\)

$$\begin{aligned} a_{1}&=5\\ d&=1.75\\ a_{n}&=20.75\\\\ a_{n}&=a_{1}+\left( n-1\right) d\\ 20.75&=5+\left( n-1\right) 1.75\\ 1.75\left( n-1\right) &=20.75-5\\ 1.75\left( n-1\right) &=15.75\\\\ \left( n-1\right) &=\dfrac{15.75}{1.75}\\\\ &=\dfrac{1575}{175}\\\\ &=9\\ n-1&=9\\ n&=10\end{aligned}$$ Ram Kali weekly saving become ₹ 20.75 in 10th week