Predicting What Comes Next: Exploring Sequences and Progressions — CBSE Class 9 Maths Important Questions

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. $a_{10}=a+(10-1)d=3+9\times4=3+36=39$a_10=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.

Here $a=2$a=2 and $d=6-2=4$d=6-2=4.

$a_{15}=a+(15-1)d=2+14\times4=2+56=58$a_15=a+(15-1)d=2+14×4=2+56=58.

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

$a_5=a+4d=19$a_5=a+4d=19 ... (1)

$a_8=a+7d=31$a_8=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.

Here $a=5,\ d=4,\ n=20$a=5, d=4, n=20.

$S_n=\dfrac{n}{2}\big(2a+(n-1)d\big)=\dfrac{20}{2}\big(2\times5+(20-1)\times4\big)$S_n=n/2(2a+(n-1)d)=20/2(2×5+(20-1)×4)

$=10\big(10+76\big)=10\times86=860$=10(10+76)=10×86=860.

Here $a=3,\ d=4$a=3, d=4. Use $S_n=\dfrac{n}{2}\big(2a+(n-1)d\big)=210$S_n=n/2(2a+(n-1)d)=210:

$\dfrac{n}{2}\big(6+(n-1)4\big)=210\Rightarrow n(4n+2)=420\Rightarrow 4n^2+2n-420=0$n/2(6+(n-1)4)=210 n(4n+2)=420 4n^2+2n-420=0.

Divide by $2$2: $2n^2+n-210=0$2n^2+n-210=0. Factorise: $(2n+21)(n-10)=0$(2n+21)(n-10)=0.

Since $n>0$n>0, $n=10$n=10. So $10$10 terms are needed.

Given $a=5,\ l=89,\ S_n=705$a=5, l=89, S_n=705.

Number of terms: using $S_n=\dfrac{n}{2}(a+l)$S_n=n/2(a+l):

$705=\dfrac{n}{2}(5+89)=\dfrac{n}{2}\times94=47n\Rightarrow n=\dfrac{705}{47}=15$705=n/2(5+89)=n/2×94=47n n=705/47=15.

Common difference: the last term is the $15$15th term, so $l=a+(n-1)d$l=a+(n-1)d:

$89=5+(15-1)d\Rightarrow 84=14d\Rightarrow d=6$89=5+(15-1)d 84=14d d=6.

Hence there are $15$15 terms and the common difference is $6$6.

The savings form an AP with $a=200$a=200 and $d=50$d=50.

(a) 12th-month saving: $a_{12}=a+(12-1)d=200+11\times50=200+550=\text{Rs }750$a_12=a+(12-1)d=200+11×50=200+550=Rs 750.

(b) Total in a year ($12$12 months): $S_{12}=\dfrac{12}{2}\big(2\times200+(12-1)\times50\big)=6\big(400+550\big)=6\times950=\text{Rs }5700$S_12=12/2(2×200+(12-1)×50)=6(400+550)=6×950=Rs 5700.

(c) Month when saving is Rs 750: set $a_n=750$a_n=750: $200+(n-1)50=750\Rightarrow(n-1)50=550\Rightarrow n-1=11\Rightarrow n=12$200+(n-1)50=750(n-1)50=550 n-1=11 n=12. So in the $12$12th month.

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) $a_{10}=20+(10-1)\times2=20+18=38$a_10=20+(10-1)×2=20+18=38 seats.

(iii) $a_{15}=20+(15-1)\times2=20+28=48$a_15=20+(15-1)×2=20+28=48 seats.

(iv) $S_{15}=\dfrac{15}{2}(a+l)=\dfrac{15}{2}(20+48)=\dfrac{15}{2}\times68=15\times34=510$S_15=15/2(a+l)=15/2(20+48)=15/2×68=15×34=510 seats.