Arithmetic Progression — ICSE Class 10 Maths Important Questions

Answer: (a) $4$4.

$d=7-3=4$d=7-3=4 (the difference between consecutive terms is constant).

Answer: (a) $29$29.

$a=2,\ d=3$a=2, d=3; $a_{10}=a+9d=2+9(3)=29$a_10=a+9d=2+9(3)=29.

Answer: (a) $16$16th.

$a=3,\ d=5$a=3, d=5; $a_n=3+(n-1)5=78\Rightarrow(n-1)5=75\Rightarrow n-1=15\Rightarrow n=16$a_n=3+(n-1)5=78(n-1)5=75 n-1=15 n=16.

Answer: (a) $6n-1$6n-1.

$a_n=S_n-S_{n-1}=(3n^2+2n)-\big[3(n-1)^2+2(n-1)\big]=3(2n-1)+2=6n-1$a_n=S_n-S_n-1=(3n^2+2n)-[3(n-1)^2+2(n-1)]=3(2n-1)+2=6n-1.

Answer: (a) With $a=5,\ d=6$a=5, d=6: $5+(n-1)6=301\Rightarrow n-1=\dfrac{296}{6}=49\tfrac13\Rightarrow n=50\tfrac13$5+(n-1)6=301 n-1=296/6=4913 n=5013. This is not a whole number, so $301$301 is not a term; R correctly explains A.

For three consecutive terms of an AP, twice the middle term equals the sum of the other two:

$2(3k+3)=(2k+1)+(5k-1)\Rightarrow 6k+6=7k\Rightarrow k=\mathbf{6}.$2(3k+3)=(2k+1)+(5k-1) 6k+6=7k k=6.

(Check: the terms are $13,21,29$13,21,29 with common difference $8$8.)

Here $a=1,\ d=2,\ n=20$a=1, d=2, n=20.

$S_{20}=\dfrac{20}{2}\big[2(1)+(20-1)(2)\big]=10\big[2+38\big]=10\times40=\mathbf{400}.$S_20=20/2[2(1)+(20-1)(2)]=10[2+38]=10×40=400.

Here $a=9,\ d=8$a=9, d=8. Using $S_n=636$S_n=636:

$\frac{n}{2}\big[2(9)+(n-1)8\big]=636\Rightarrow\frac{n}{2}(8n+10)=636.$n/2[2(9)+(n-1)8]=636/2(8n+10)=636.

$n(4n+5)=636\Rightarrow 4n^2+5n-636=0.$n(4n+5)=636 4n^2+5n-636=0.

$n=\dfrac{-5\pm\sqrt{25+10176}}{8}=\dfrac{-5\pm101}{8}$n=-5±√25+10176/8=-5±101/8.

Taking the positive value, $n=\dfrac{96}{8}=\mathbf{12}$n=96/8=12.

$a+3d=11$a+3d=11 ...(1) and $a+7d=23$a+7d=23 ...(2).

Subtracting (1) from (2): $4d=12\Rightarrow d=3$4d=12 d=3; then $a=11-9=2$a=11-9=2.

The AP is $\mathbf{2,5,8,11,\dots}$2,5,8,11,

$a_{20}=a+19d=2+19(3)=\mathbf{59}.$a_20=a+19d=2+19(3)=59.

Let the numbers be $a-d,\ a,\ a+d$a-d, a, a+d.

Sum: $(a-d)+a+(a+d)=3a=24\Rightarrow a=8$(a-d)+a+(a+d)=3a=24 a=8.

Product: $(8-d)(8)(8+d)=440\Rightarrow 8(64-d^2)=440\Rightarrow 64-d^2=55\Rightarrow d^2=9\Rightarrow d=\pm3$(8-d)(8)(8+d)=440 8(64-d^2)=440 64-d^2=55 d^2=9 d=±3.

The numbers are $\mathbf{5,\ 8,\ 11}$5, 8, 11 (or $11,8,5$11,8,5).

The numbers are $14,21,28,\dots,98$14,21,28,,98: an AP with $a=14$a=14, $d=7$d=7 and last term $l=98$l=98.

Number of terms: $14+(n-1)7=98\Rightarrow n-1=12\Rightarrow n=13$14+(n-1)7=98 n-1=12 n=13.

$S_{13}=\dfrac{13}{2}(14+98)=13\times56=\mathbf{728}.$S_13=13/2(14+98)=13×56=728.

Using $S_7=49$S_7=49: $\dfrac{7}{2}(2a+6d)=49\Rightarrow 2a+6d=14\Rightarrow a+3d=7.$7/2(2a+6d)=49 2a+6d=14 a+3d=7. ...(1)

Using $S_{17}=289$S_17=289: $\dfrac{17}{2}(2a+16d)=289\Rightarrow 2a+16d=34\Rightarrow a+8d=17.$17/2(2a+16d)=289 2a+16d=34 a+8d=17. ...(2)

Subtract (1) from (2): $5d=10\Rightarrow d=2$5d=10 d=2; then $a=7-6=1$a=7-6=1.

$S_n=\frac{n}{2}\big[2(1)+(n-1)2\big]=\frac{n}{2}(2n)=\mathbf{n^2}.$S_n=n/2[2(1)+(n-1)2]=n/2(2n)=n^2.

The prizes form an AP with $n=7$n=7, $d=-20$d=-20 and sum $S_7=700$S_7=700.

$S_7=\frac{7}{2}\big[2a+(7-1)(-20)\big]=700.$S_7=7/2[2a+(7-1)(-20)]=700.

$\dfrac{7}{2}(2a-120)=700\Rightarrow 2a-120=200\Rightarrow 2a=320\Rightarrow a=160.$7/2(2a-120)=700 2a-120=200 2a=320 a=160.

Starting at ₹$160$160 and decreasing by ₹$20$20, the prizes are:

$\mathbf{₹160,\ ₹140,\ ₹120,\ ₹100,\ ₹80,\ ₹60,\ ₹40.}$₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40.

(Check: their sum $=700$=700.)

(i) The seats are $20,22,24,\dots$20,22,24,, so the common difference is $d=\mathbf{2}$d=2 (with $a=20$a=20).

(ii) $a_{15}=a+14d=20+14(2)=20+28=\mathbf{48}$a_15=a+14d=20+14(2)=20+28=48 seats.

(iii) $S_{15}=\dfrac{15}{2}\big[2(20)+(15-1)(2)\big]=\dfrac{15}{2}(40+28)=\dfrac{15}{2}(68)=15\times34=\mathbf{510}$S_15=15/2[2(20)+(15-1)(2)]=15/2(40+28)=15/2(68)=15×34=510 seats.

(iv) The $5$5th row from the back of $15$15 rows is row $15-5+1=11$15-5+1=11. $a_{11}=20+10(2)=\mathbf{40}$a_11=20+10(2)=40 seats.