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 $15625 .$