Exercise 5.2 — Solutions 

1. Fill in the blanks (use $a_n=a+(n-1)d$

(i) $a=7,d=3,n=8$

$$a_n=7+(8-1)\cdot 3=7+21=28.$$

(ii) $a=-18,n=10,a_n=0$

$$0=-18+(10-1)d\Rightarrow 9d=18\Rightarrow d=2.$$

(iii) $d=-3,n=18,a_n=-5.$ Find $a$.

$$a+(18-1)(-3)=-5\Rightarrow a-51=-5\Rightarrow a=46.$$

(iv) $a=-18.9,d=2.5,a_n=\mathrm{3.6.}$ Find $n$.

$$-18.9+(n-1)\cdot 2.5=3.6\Rightarrow (n-1)\cdot 2.5=22.5\Rightarrow n-1=9\Rightarrow n=10.$$

(v) $a=3.5,d=0,n=105.$

$$a_n=3.5(\text{every term is }3.5)\mathrm{.}$$

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2. Multiple choice — pick correct option and justify

(i) AP: $10,7,4,\dots$ so $a=10,d=-3$

$$a_{30}=10+(30-1)(-3)=10-87=-77.$$

Answer: (C) $\mathrm{}$.

(ii) AP: $-3,-\frac{1}{2},2,\dots$ Here $a=-3$, difference $d=2.5$

$$a_{11}=-3+10\cdot 2.5=-3+25=22.$$

Answer: (B) $22$.

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3. Find missing terms

I solved each box assuming a standard interpretation of the printed problem. One line in the PDF was slightly unclear; where I made an assumption I’ve noted it.

(i) $2,\boxed{{\displaystyle 14}},26$
$\boxed{{\displaystyle \text{middle}}}={\displaystyle \frac{2+26}{2}}=14$

(ii) $\boxed{{\displaystyle 18}},13,\boxed{{\displaystyle 8}},3$
Let $d=-5$ (since $13-18=-5$); then next terms are $13-5=8,8-5=3$

(iii) (ambiguous in print) — the problem shows $5,\mathrm{\_},\mathrm{\_},\mathrm{\_}$with a final term If the intended four-term AP is $5,?,?,\frac{1}{9}$, then

$$d=\frac{\frac{1}{9}-5}{3}=\frac{-44\mathrm{/}9}{3}=-\frac{44}{27},$$

so

$$a_2=5-\frac{44}{27}=\frac{91}{27},a_3=\frac{47}{27}\mathrm{.}$$

(iv) $-4,\boxed{{\displaystyle -2}},\boxed{{\displaystyle 0}},\boxed{{\displaystyle 2}},\boxed{{\displaystyle 4}},\mathrm{6 }$

Here $a_1=-4,a_6=6$ so $d=(6-(-4))\mathrm{/}5=2$

(v) $\boxed{{\displaystyle 53}},38,23,8,-7,-22$

From $a_2=38$ and $a_6=-22$ we get $4d=-60\Rightarrow d=-15$, so $a_1=38-d=53$

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4. Which term of $3,8,13,18,\dots$ is $78$?

$a=3,d=5$

$$3+(n-1)5=78\Rightarrow 5(n-1)=75\Rightarrow n-1=15\Rightarrow n=16.$$

Answer: 16th term.

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5. Number of terms

(i) $7,13,19,\dots ,205.$ $a=7,d=6$

$$7+(n-1)6=205\Rightarrow 6(n-1)=198\Rightarrow n=34.$$

(ii) $18,15.5,13,\dots ,-47$ (Interpretation: second term is $15\frac{1}{2}=15.5$)
Then $d=15.5-18=-2.5$

Solve

$$18+(n-1)(-2.5)=-47\Rightarrow (n-1)=\frac{-65}{-2.5}=26\Rightarrow n=27.$$

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6. Is $-150$ a term of $11,8,5,2,\dots$?

Here $a=11,d=-3$. Solve

$$11+(n-1)(-3)=-150\Rightarrow (n-1)=\frac{-161}{-3}=53\frac{2}{3},$$

not an integer.
Answer: No, $-150$ is not a term.

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7. $a_{11}=38,a_{16}=73$. Find $a_{31}$.

Let $a+10d=38$ and $a+15d=73$. Subtract: $5d=35\Rightarrow d=7$

Then $a=38-70=-32$

$$a_{31}=a+30d=-32+30\cdot 7=-32+210=178.$$

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8. AP of 50 terms; $a_3=12,a_{50}=106$. Find $a_{29}$.

From $a+2d=12\Rightarrow a=12-2d$. Also $a+49d=106$. Substitute:

$$12-2d+49d=106\Rightarrow 47d=94\Rightarrow d=2,a=12-4=8.$$

So $a_{29}=8+28\cdot 2=8+56=64$

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9. $a_3=4,a_9=-8$. Which term is zero?

$a+2d=4,a+8d=-8\Rightarrow 6d=-12\Rightarrow d=-2$

$\mathrm{<span\; class="katex"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mspace"> </span></span></span></span><span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">Then </span>}$$a=4-2d=8$. Solve

$$8+(n-1)(-2)=0\Rightarrow 8-2(n-1)=0\Rightarrow n=5.$$

Answer: 5th term is zero.

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10. $a_{17}$ exceeds $a_{10}$ by $7$. Find $d$.

$$(a+16d)-(a+9d)=7\Rightarrow 7d=7\Rightarrow d=1.$$

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11. In AP $3,15,27,39,\dots$ which term is $132$ more than its $54$th term?

Here $a=3,d=12$.

$a_{54}=3+53\cdot 12=\mathrm{639<span\; class="katex"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mord">.</span></span></span></span><span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">\; Need </span>}$$a_k=639+132=771$. Solve

$$3+(k-1)12=771\Rightarrow 12(k-1)=768\Rightarrow k-1=64\Rightarrow k=65.$$

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12. Two APs have same $d$. Difference between their 100th terms is $100$. What is difference between their 1000th terms?

If APs are $a+(n-1)d$ and $b+(n-1)d$, difference at any $n$ is $a-b$. Since difference at $n=100$ is $100$, the difference at $n=1000$ is also

$$\boxed{{\displaystyle 100}}\mathrm{.}$$

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

Smallest three-digit divisible by 7: $105$. Largest ≤999 divisible by 7: $994$. Count:

$$\frac{994-105}{7}+1=\frac{889}{7}+1=127+1=128.$$

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

Smallest multiple $>10$ is $12$; largest $\le 250$ is $248$. Count:

$$\frac{248-12}{4}+1=\frac{236}{4}+1=59+1=60.$$

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15. For what value of $n$ are the $n$-th terms equal for the APs $63,65,67,\dots$ and $3,10,17,\dots$?

First AP: $a_1=63,d_1=2$ so $a_n^{(1)}=63+2(n-1)$.
Second AP: $a_1^{\mathrm{\prime }}=3,d_2=7$ so $a_n^{(2)}=3+7(n-1)$.

Set equal:

$$63+2(n-1)=3+7(n-1)\Rightarrow 61+2n=-4+7n\Rightarrow 5n=65\Rightarrow n=13.$$

Q16. A sum of ₹700 is to be used to give seven cash prizes. Each prize is ₹20 less than the preceding prize. Find all seven prizes.

Let the prizes form an AP with first term $a$, common difference $d=-20$, number of terms $n=7$. The sum is

$$S_n=\frac{n}{2}(2a+(n-1)d)=700.$$

Substitute $n=7$, $d=-20$:

$$700=\frac{7}{2}(2a+6(-20))\Rightarrow 700=\frac{7}{2}(2a-120)\mathrm{.}$$

Multiply by $2$: $1400=7(2a-120)\Rightarrow 200=2a-120\Rightarrow 2a=320\Rightarrow a=160.$

Thus the seven prizes are

$$160,140,120,100,80,60,40(\text{in ₹})\mathrm{.}$$

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Q17. Each section of Class I plants 1 tree, Class II plants 2 trees, …, Class XII plants 12 trees. There are 3 sections in each class. How many trees in total?

Total trees $=3\times (1+2+\cdots +12)=3\cdot \frac{12\cdot 13}{2}=3\cdot 78=234.$

So 234 trees will be planted.

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Q18. A spiral is made of 13 successive semicircles of radii $0.5,1.0,1.5,\dots$ Find the total length. (Take $\pi =\frac{22}{7}$.)

Length of a semicircle of radius $r$ is $\frac{1}{2}(2\pi r)=\pi r$. So total length

$$L=\pi \sum _{k=1}^{13}{r}_k,r_k=0.5+(k-1)\cdot \mathrm{0.5.}$$

The radii form an AP with $a=0.5,d=0.5,n=13$. Sum of radii:

$$\sum r_k=\frac{13}{2}(2\cdot 0.5+(13-1)\cdot 0.5)=\frac{13}{2}(1+6)=\frac{13}{2}\cdot 7=\frac{91}{2}\mathrm{.}$$

Thus

$$L=\pi \cdot \frac{91}{2}=\frac{22}{7}\cdot \frac{91}{2}=\frac{22\times 91}{14}=\frac{2002}{14}=143\text{ cm}\mathrm{.}$$

Total length = 143 cm.

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Q19. 200 logs are stacked: bottom row 20 logs, next 19, next 18, … . In how many rows are the 200 logs placed and how many logs in the top row?

This is a finite AP with $a=20,d=-1$. Let number of rows be $n$ and last row have $l=a+(n-1)d$. Sum:

$$S_n=\frac{n}{2}(a+l)=200.$$

Using $l=20-(n-1)=21-n$, we get

$$\frac{n}{2}(20+(21-n))=\frac{n}{2}(41-n)=200\Rightarrow n(41-n)=400.$$

So $n^2-41n+400=0$. Discriminant $\mathrm{\Delta }={41}^2-4\cdot 400=1681-1600=81$, $\sqrt{\mathrm{\Delta }}=9$

$$n=\frac{41\pm 9}{2}=\{25,16\}\mathrm{.}$$

Physically the number of logs in the top row must be non-negative: for $n=25$ the top row would be $20-(25-1)=-4$(impossible). So take $n=16$. Top row has

$$l=20-(16-1)=20-15=5\text{ logs}\mathrm{.}$$

Answer: 16 rows, top row has 5 logs.

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Q20. Potato race: bucket is at start, first potato is 5 m from bucket, others 3 m apart; 10 potatoes. Competitor picks each potato and returns it to bucket, repeats. Total distance?

Let distances of potatoes from bucket be an AP: $a=5,d=3,n=10$. Sum of distances to each potato:

$$\sum _{i=1}^{10}{d}_i=\frac{10}{2}(2\cdot 5+(10-1)\cdot 3)=5(10+27)=5\cdot 37=185\text{ m}\mathrm{.}$$

For each potato the competitor runs to it and back, so total distance $=2\times$ (sum of distances) $=2\times 185=370$

Total distance = 370 metres.