Sequences and Progressions Class 9 CBSE Important Questions with Answers

Answer: (a) $14$14.

Each term increases by $3$3 (common difference $d=3$d=3), so $11+3=14$11+3=14.

Answer: (b) $-3$-3.

$d=\text{any term}-\text{previous term}=4-7=-3$d=any term-previous term=4-7=-3 (the sequence decreases by $3$3 each step).

Answer: (a) $39$39.

Here $a=3,\ d=4$a=3, d=4. $a_{10}=a+(10-1)d=3+9\times4=3+36=39$a₁₀=a+(10-1)d=3+9×4=3+36=39.

Answer: (b) $19$19th.

$a=5,\ d=3$a=5, d=3. Set $a_n=59$a_n=59: $5+(n-1)3=59\Rightarrow(n-1)3=54\Rightarrow n-1=18\Rightarrow n=19$5+(n-1)3=59⇒(n-1)3=54⇒ n-1=18⇒ n=19.

Answer: (d) A is false but R is true. In $2,4,8,16,\ldots$2,4,8,16,… the differences are $2,4,8,\ldots$2,4,8,… (not constant), so it is not an AP. R correctly states the defining property of an AP.

Start at $6$6 and keep adding $d=-2$d=-2:

$6,\ 6+(-2)=4,\ 4+(-2)=2,\ 2+(-2)=0$6, 6+(-2)=4, 4+(-2)=2, 2+(-2)=0.

The first four terms are $6,\ 4,\ 2,\ 0$6, 4, 2, 0.

Stages $0,1,2,3$0,1,2,3 have $1,\ 3,\ 9,\ 27$1, 3, 9, 27 shaded triangles. Each term is $3$3 times the previous one, so this is a geometric progression (GP) with first term $1$1 and common ratio $r=3$r=3.

At stage $n$n there are $3^n$3^n shaded triangles, so at stage $5$5 there are $3^5=243$3⁵=243.

Using $a_n=a+(n-1)d$a_n=a+(n-1)d:

$a_5=a+4d=19$a₅=a+4d=19 ... (1)

$a_8=a+7d=31$a₈=a+7d=31 ... (2)

Subtract (1) from (2): $3d=12\Rightarrow d=4$3d=12⇒ d=4.

Put $d=4$d=4 in (1): $a+16=19\Rightarrow a=3$a+16=19⇒ a=3.

So the first term is $3$3 and the common difference is $4$4.

Write the sum forwards and backwards and add the two lines term by term:
$\begin{aligned}S&=1+2+3+\dots+100\\S&=100+99+98+\dots+1\\2S&=101+101+101+\dots+101\ (100\text{ times})\end{aligned}$S&=1+2+3+…+100; S&=100+99+98+…+1; 2S&=101+101+101+…+101 (100 times)
So $2S=100\times101$2S=100×101 and $S=\dfrac{100\times101}{2}=5050$S=100×101/2=5050.

In the same way, for the first $n$n natural numbers $1+2+\dots+n=\dfrac{n(n+1)}{2}$1+2+…+n=(n(n+1))/2.

For $n=50$n=50: $\dfrac{50\times51}{2}=1275$50×51/2=1275.

(a) Explicit rule (the $n$nth term directly): $a_n=3n+2$a_n=3n+2. Check: $a_1=5,\ a_4=14$a₁=5, a₄=14.
Recursive rule (each term from the one before): $a_1=5$a₁=5 and $a_n=a_{n-1}+3$a_n=a_n-1+3 for $n\ge2$n≥2.

(b) Each term is the sum of the two terms before it: $a_1=a_2=1$a₁=a₂=1 and $a_n=a_{n-1}+a_{n-2}$a_n=a_n-1+a_n-2 for $n\ge3$n≥3.
Next three terms: $5+8=13$5+8=13, $8+13=21$8+13=21, $13+21=34$13+21=34.
It is not an AP: the differences $0,1,1,2,3,\ldots$0,1,1,2,3,… are not constant.

(a) $\dfrac{6}{3}=\dfrac{12}{6}=\dfrac{24}{12}=2$6/3=12/6=24/12=2. Each term is the previous term multiplied by the same number, so it is a GP with common ratio $r=2$r=2 (first term $a=3$a=3).

(b) $a_n=ar^{\,n-1}=3\times2^{\,n-1}$a_n=ar^ n-1=3×2^ n-1.

(c) $a_8=3\times2^7=3\times128=384$a₈=3×2⁷=3×128=384.

(d) $3\times2^{\,n-1}=768\Rightarrow2^{\,n-1}=256=2^8\Rightarrow n-1=8\Rightarrow n=9$3×2^ n-1=768⇒2^ n-1=256=2⁸⇒ n-1=8⇒ n=9. So $768$768 is the $9$9th term.

(a) Plan A adds a fixed amount, so it is an AP with $a=200$a=200, $d=50$d=50: $A_n=200+(n-1)50=150+50n$A_n=200+(n-1)50=150+50n.
Plan B multiplies by a fixed number, so it is a GP with $a=5$a=5, $r=2$r=2: $B_n=5\times2^{\,n-1}$B_n=5×2^ n-1.

(b) $A_{12}=150+600=\text{Rs }750$A₁₂=150+600=Rs 750; $B_{12}=5\times2^{11}=5\times2048=\text{Rs }10240$B₁₂=5×2¹¹=5×2048=Rs 10240.

(c) Compare month by month:

Plan B first saves more in the 8th month ($640>550$640>550). A GP that grows by a ratio greater than $1$1 eventually overtakes any AP.

The seats per row form an AP with $a=20$a=20 and $d=2$d=2.

(i) Common difference $d=2$d=2.

(ii) $a_{10}=20+(10-1)\times2=20+18=38$a₁₀=20+(10-1)×2=20+18=38 seats.

(iii) $a_{15}=20+(15-1)\times2=20+28=48$a₁₅=20+(15-1)×2=20+28=48 seats.

(iv) Explicit: $a_n=20+(n-1)\times2=2n+18$a_n=20+(n-1)×2=2n+18. Recursive: $a_1=20$a₁=20 and $a_n=a_{n-1}+2$a_n=a_n-1+2 for $n\ge2$n≥2.