Sequences and Series

Class 11 NCERT

NCERT

1   Write the first five terms of the sequences whose $n^{th}$ term is $ a_n=n(n+2).$

Solution :

$a_n=n(n+2)$$\\$ Substituating $n=1,2,3,4$ and $5,$ we obtain$\\$ $ a_1=1(1+2)=3\\ a_2=2(2+2)=8\\ a_3=3(3+2)=15\\ a_4=4(4+2)=24\\ a_5=5(5+2)=35$$\\$ Therefore, the required terms are $3, 8, 15, 24$ and $35.$

2   Write the first five terms of the sequences whose $n^{th}$ term is $ a_n=\dfrac{n}{n+1}$

Solution :

$ a_n=\dfrac{n}{n+1}$$\\$ Substituting $ n=1,2,3,4,5,$ we obtain$\\$ $a_1=\dfrac{1}{1+1}=\dfrac{1}{2},a_2=\dfrac{2}{2+1}=\dfrac{2}{3},\\ a_3=\dfrac{3}{3+1}=\dfrac{3}{4},a_4=\dfrac{4}{4+1}=\dfrac{4}{5},\\ a_5=\dfrac{5}{5+1}=\dfrac{5}{6}$$\\$ Therefore, the required terms are $\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5}$ and $\dfrac{5}{6}$

3   Write the first five terms of the sequences whose $n^{th}$ term is $ a_n=2^n$

Solution :

$a_n=2^n$$\\$ Substituting $n=1,2,3,4,5,$ we obtain$\\$ $a_1=2^1=2\\ a_2=2^2=4\\ a_3=2^3=8\\ a_4=2^4=16\\ a_5=2^5=32$$\\$ Therefore, the required terms are $2, 4, 8, 16$ and $32.$

4   Write the first five terms of the sequences whose $n^{th}$ term is $ a_n=\dfrac{2n-3}{6}$

Solution :

Substituting $n = 1,2,3,4,5 ,$ we obtain$\\$ $a_1=\dfrac{2*1-3}{6}=\dfrac{-1}{6}\\ a_2=\dfrac{2*2-3}{6}=\dfrac{1}{6}\\ a_3=\dfrac{2*3-3}{6}=\dfrac{3}{6}=\dfrac{1}{2}\\ a_4=\dfrac{2*4-3}{6}=\dfrac{5}{6}\\ a_5=\dfrac{2*5-3}{6}=\dfrac{7}{6}$$\\$ Therefore, the required terms are $ \dfrac{-1}{6},\dfrac{1}{6},\dfrac{1}{2},\dfrac{5}{6}$ and $ \dfrac{7}{6}.$

5   Write the first five terms of the sequences whose $n^{th}$ term is $ a_n=(-1)^{n-1}5^{n+1}$

Solution :

Substituting $n = 1,2,3,4,5 ,$ we obtain $\\$ $a_1=(-1)^{1-1}5^{1+1}=5^2=25\\ a_2=(-1)^{2-1}5^{2+1}=-5^3=-125\\ a_3=(-1)^{3-1}5^{3+1}=5^4=625\\ a_4=(-1)^{4-1}5^{4+1}=-5^5=-3125\\ a_5=(-1)^{5-1}5^{5+1}= 5^6=15625$$\\$ Therefore, the required terms are $25, - 125,625, - 3125$ and $15625 .$

6   Write the first five terms of the sequences whose $n^{th} $ term is $ a_n=n\dfrac{n^2+5}{4}$

Solution :

Substituting $n=1,2,3,4,5,$ we obtain$\\$ $ a_1=1.\dfrac{1^2+5}{4}=\dfrac{6}{4}=\dfrac{3}{2}\\ a_2=2.\dfrac{2^2+5}{4}=2.\dfrac{9}{4}=\dfrac{9}{2}\\ a_3=3.\dfrac{3^2+5}{4}=3.\dfrac{14}{4}=\dfrac{21}{2}\\ a_4=4.\dfrac{4^2+5}{4}=21\\ a_5=5.\dfrac{5^2+5}{4}=5.\dfrac{30}{4}=\dfrac{75}{2}$$\\$ Therefore, the required terms are $ \dfrac{3}{2},\dfrac{9}{2},\dfrac{21}{2},21 $ and $ \dfrac{75}{2}$

7   Find the $17^{ th}$ and $24^{ th}$ term in the following sequence whose $n^{ th}$ term is $a _n = 4 _n - 3$

Solution :

Substituting $n=17,$ we obtain$\\$ $a_17=4(17)-3=68-3=65$$\\$ Substituting $ n=24, $ we obtain $\\$ $a_{24}=4(24)-3=96-3=93$

8   Find the $7^{th}$ term in the following sequence whose $n^{th}$ term is $a_ n=\dfrac{n^2}{2^n}$

Solution :

Substituting $n=7,$ we obtain$\\$ $ a_7=\dfrac{7^2}{2^7}=\dfrac{49}{128}$

9   Find the $9^{ th }$ term in the following sequence whose $n ^{th}$ term is $ a_ n=(-1)^{n-1} n^3$

Solution :

Substituting $n = 7$ , we obtain$\\$ $a_9=(-1)^{9-1}(9)^3=(9)^3=729$

10   Find the $20^{ th}$ term in the following sequence whose $n^{ th}$ term is $a_ n=\dfrac{n(n-2)}{n+3}$

Solution :

Substituting $n = 20$, we obtain $\\$ $a_{20}=\dfrac{20(20-2)}{20+3}=\dfrac{20(18)}{23} =\dfrac{360}{23}$