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Exercise 5.2 · Q2

Q.Write the first 6 terms of the sequences whose nthn^{th} term ana_n is given below.

(i) an={n+1if n is oddnif n is evena_n=\begin{cases} n+1 & \text{if } n \text{ is odd} \\ n & \text{if } n \text{ is even}\end{cases}
(ii) an={1if n=12if n=2an−1+an−2if n>2a_n=\begin{cases} 1 & \text{if } n=1 \\ 2 & \text{if } n=2 \\ a_{n-1}+a_{n-2} & \text{if } n>2\end{cases}
(iii) an={nif n is 1,2 or 3an−1+an−2+an−3if n>3a_n=\begin{cases} n & \text{if } n \text{ is } 1, 2 \text{ or } 3 \\ a_{n-1}+a_{n-2}+a_{n-3} & \text{if } n>3\end{cases}
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✓ Free question

Substitute directly for the piecewise rule (i), and build up term-by-term for the two recursively-defined sequences (ii),(iii).

Step 1. (i) an=n+1a_n=n+1 (odd nn), an=na_n=n (even nn).

n=1n=1(odd):2:2; n=2n=2(even):2:2; n=3n=3(odd):4:4; n=4n=4(even):4:4; n=5n=5(odd):6:6; n=6n=6(even):6:6.

Sequence: 2,2,4,4,6,62,2,4,4,6,6.

Step 2. (ii) a1=1,a2=2,an=an−1+an−2a_1=1,a_2=2,a_n=a_{n-1}+a_{n-2} for n>2n>2.

a3=a2+a1=2+1=3a_3=a_2+a_1=2+1=3; a4=a3+a2=3+2=5a_4=a_3+a_2=3+2=5; a5=a4+a3=5+3=8a_5=a_4+a_3=5+3=8; a6=a5+a4=8+5=13a_6=a_5+a_4=8+5=13.

Sequence: 1,2,3,5,8,131,2,3,5,8,13.

Step 3. (iii) an=na_n=n for n=1,2,3n=1,2,3; an=an−1+an−2+an−3a_n=a_{n-1}+a_{n-2}+a_{n-3} for n>3n>3.

a1=1,a2=2,a3=3a_1=1,a_2=2,a_3=3. a4=a3+a2+a1=3+2+1=6a_4=a_3+a_2+a_1=3+2+1=6; a5=a4+a3+a2=6+3+2=11a_5=a_4+a_3+a_2=6+3+2=11; a6=a5+a4+a3=11+6+3=20a_6=a_5+a_4+a_3=11+6+3=20.

Sequence: 1,2,3,6,11,201,2,3,6,11,20.

✓Final answer

(i) 2,2,4,4,6,62,2,4,4,6,6; (ii) 1,2,3,5,8,131,2,3,5,8,13; (iii) 1,2,3,6,11,201,2,3,6,11,20.

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