Exercise 5.3 — Solutions (Questions 1–20)

Useful formulas

$$a_n=a+(n-1)d,S_n=\frac{n}{2}(2a+(n-1)d)=\frac{n}{2}(a+l)\mathrm{.}$$

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1. Find the sum of the following APs

(i) $2,7,12,\dots$ (10 terms).
$a=2,d=5,n=10$

$$S_{10}=\frac{10}{2}(2\cdot 2+9\cdot 5)=5(4+45)=5\cdot 49=245.$$

(ii) $-37,-33,-29,\dots$(12 terms).
$a=-37,d=4,n=12$

$$S_{12}=\frac{12}{2}(2(-37)+11\cdot 4)=6(-74+44)=6(-30)=-180.$$

(iii) $0.6,1.7,2.8,\dots$ (100 terms).
$a=0.6,d=1.1,n=100$

$$S_{100}=\frac{100}{2}(2\cdot 0.6+99\cdot 1.1)=50(1.2+108.9)=50\cdot 110.1=5505.$$

(iv)

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2. Find the sums below

(i)

(ii) $34+32+30+\cdots +10$
$a=34,d=-2,l=10.$ Number of terms:

$$n=\frac{10-34}{-2}+1=13.$$

$$S_{13}=\frac{13}{2}(34+10)=\frac{13}{2}\cdot 44=13\cdot 22=286.$$

(iii) $-5+(-8)+(-11)+\cdots +(-230)$
$a=-5,d=-3,l=-230$ Number of terms:

$$n=\frac{-230-(-5)}{-3}+1=\frac{-225}{-3}+1=76.$$
$$S_{76}=\frac{76}{2}(-5-230)=38(-235)=-8930.$$

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3. Short AP problems (solved stepwise)

(i) $a=5,d=3,a_n=50$
$50=5+(n-1)3\Rightarrow n=16$

$$S_{16}=\frac{16}{2}[2\cdot 5+15\cdot 3]=8(10+45)=8\cdot 55=440.$$

(ii) $a=7,a_{13}=35$
$7+12d=35\Rightarrow d={\displaystyle \frac{7}{3}}\mathrm{.}$

$$S_{13}=\frac{13}{2}(14+12\cdot \frac{7}{3})=273.$$

(iii) $a_{12}=37,d=3\Rightarrow a=4$

$$S_{12}=6(8+33)=246.$$

(iv) $a_3=15,S_{10}=125$
From $a+2d=15$ and $5(2a+9d)=125$ we get $d=-1,a=17,a_{10}=8$

(v) $d=5,S_9=75$
$\frac{9}{2}(2a+40)=75\Rightarrow a=-\frac{35}{3},a_9=\frac{85}{3}\mathrm{.}$

(vi) $a=2,d=8,S_n=90$
Solve $n(4n-2)=90\Rightarrow n=5,a_5=34$

(vii) $a=8,a_n=62,S_n=210$.
$\frac{n}{2}(8+62)=210\Rightarrow n=6,d=\frac{54}{5}\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$

(viii) $a_n=4,d=2,S_n=-14$
From $a=6-2n$ and $S_n=n(5-n)=-14\Rightarrow n=7,a=-8$

(ix) $a=3,n=8,S=192$

$$192=4(6+7d)\Rightarrow d=6.$$

(x) $l=28,S=144,n=9$

$$144=\frac{9}{2}(a+28)\Rightarrow a=4.$$

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

$a=3,d=5$. Solve $3+5(n-1)=78\Rightarrow n=16.$

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5. Number of terms

(i) $7,13,19,\dots ,205$. $n={\displaystyle \frac{205-7}{6}}+1=34.$

(ii) $18,15.5,13,\dots ,-\mathrm{47,}$ $d=-2.5\Rightarrow n={\displaystyle \frac{-47-18}{-2.5}}+1=\mathrm{27<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$

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

$a=11,d=-3$. Solve $11+(n-1)(-3)=-150\Rightarrow n$ not integer ⇒ No.

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

$5d=35\Rightarrow d=7,a=-32.$

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

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

From $a+2d=12,a+49d=106\Rightarrow d=2,a=8$

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

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

$6d=-12\Rightarrow d=-2,a=8$. Solve $8-2(n-1)=0\Rightarrow n=5$

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

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

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

Here $a=3,d=12$. $a_{54}=639$. Need $a_k=771\Rightarrow k=65$

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

Difference is constant $=100$.

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

Smallest $=105$, largest $=994$. Count $={\displaystyle \frac{994-105}{7}}+1=128.$

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

Smallest $=12$, largest $=248$. Count $={\displaystyle \frac{248-12}{4}}+1=60.$

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15. For what $n$ do the $n$-th terms of $63,65,67,\dots$ and $3,10,17,\dots$ coincide?

Solve $63+2(n-1)=3+7(n-1)\Rightarrow n=13.$

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16. ₹700 used for seven cash prizes each ₹20 less than the preceding. Find prizes.

$n=7,d=-20,S_7=700\Rightarrow a=160.$

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

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17. Class I–XII: each section of Class $k$ plants $k$ trees; 3 sections in each class. Total trees:

$$3\sum _{k=1}^{12}k=3\cdot \frac{12\cdot 13}{2}=234.$$

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18. Spiral of 13 semicircles with radii $0.5,1.0,1.5,\dots$ Total length $(\pi =\frac{22}{7})$.

Semicircle length = $\pi r$. Radii sum:

$$\sum r_k=\frac{13}{2}[2\cdot 0.5+12\cdot 0.5]=\frac{91}{2}\mathrm{.}$$
$$L=\pi \cdot \frac{91}{2}=\frac{22}{7}\cdot \frac{91}{2}=143\text{ cm}\mathrm{.}$$

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19. 200 logs stacked: rows 20,19,18,… How many rows and top row?

Let $a=20,d=-1$. Solve ${\displaystyle \frac{n}{2}}(41-n)=200\Rightarrow n^2-41n+400=0$

 $n=25,16$. Physical solution $n=16$. Top row = $20-(16-1)=5$ logs.

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20. Potato race: potatoes at distances $5,8,11,\dots$ (10 potatoes). Competitor runs to each and back. Total distance?

Sum of distances $={\displaystyle \frac{10}{2}}[2\cdot 5+9\cdot 3]=185$. Round trips ⇒ $2\times 185=370$