# Find the next five terms of each of the following sequences given by:

Question:

Find the next five terms of each of the following sequences given by:

(i) $a_{1}=1, a_{n}=a_{n-1}+2, n \geq 2$

(ii) $a_{1}=a_{2}=2, a_{n}=a_{n-1}-3, n>2$

(iii) $a_{1}=-1, a_{n}=\frac{a_{n}-1}{n}, n \geq 2$

(iv) $a_{1}=4, a_{n}=4 a_{n-1}+3, n>1$.

Solution:

In the given problem, we are given the first, second term and the nth term of an A.P.

We need to find its next five terms

(i) $a_{1}=1, a_{n}=a_{n-1}+2, n \geq 2$

Here, we are given that $n \geq 2$

So, the next five terms of this A.P would be $a_{2}, a_{3}, a_{4}, a_{5}$ and $a_{6}$

Now $a_{1}=1$ ..........(1)

So, to find theterm we use, we get,

$a_{2}=a_{2-1}+2$

$a_{2}=a_{1}+2$

$a_{2}=1+2$ (Using 1)

$a_{2}=3 \ldots \ldots$ (2)

For, using, we get,

$a_{3}=a_{3-1}+2$

$a_{3}=a_{2}+2$

$a_{3}=3+2($ Using 2$)$

$a_{3}=5 \ldots \ldots(3)$

For, using, we get,

$a_{4}=a_{4-1}+2$

$a_{4}=a_{3}+2$

$a_{4}=5+2$ (Using 3)

$a_{4}=7$.....(4)

For, using, we get,

$a_{5}=a_{5-1}+2$

$a_{5}=a_{4}+2$

$a_{5}=7+2($ Using 4$)$

$a_{5}=9$.........(5)

For, using, we get,

$a_{6}=a_{6-1}+2$

$a_{6}=a_{5}+2$

$a_{6}=9+2$ (Using 5)

$a_{6}=11$

Therefore, the next five terms, of the given A.P are

$a_{2}=3, a_{3}=5, a_{4}=7, a_{5}=9, a_{6}=11$

(ii) $a_{1}=a_{2}=2, a_{n}=a_{n-1}-3, n>2$

Here, we are given that $n>2$

So, the next five terms of this A.P would be $a_{3}, a_{4}, a_{5}, a_{6}$ and $a_{7}$

Now $a_{1}=a_{2}=2$........(1)

So, to find the $a_{3}$ term we use $n=3$, we get,

$a_{3}=a_{3-1}-3$

$a_{3}=a_{2}-3$

$a_{3}=2-3$ (Using 1 )

$a_{3}=-1 \ldots \ldots(2)$

For $a_{4}$, using $n=4$, we get,

$a_{4}=a_{4-1}-3$

$a_{4}=a_{3}-3$

$a_{4}=-1-3$ (Using 2 )

$a_{4}=-4 \ldots \ldots(3)$

For $a_{5}$, using $n=5$, we get,

$a_{5}=a_{5-1}-3$

$a_{5}=a_{4}-3$

$a_{5}=-4-3$ (Using 3)

$a_{5}=-7$........(4)

For $a_{6}$, using $n=6$, we get,

$a_{6}=a_{6-1}-3$

$a_{6}=a_{5}-3$

$a_{6}=-7-3($ Using 4)

$a_{6}=-10 \ldots \ldots(5)$

For $a_{7}$, using $n=7$, we get,

$a_{7}=a_{7-4}-3$

$a_{7}=a_{6}-3$

$a_{7}=-10-3($ Using 5$)$

$a_{7}=-13$

Therefore, the next five terms, of the given A.P are

$a_{3}=-1, a_{4}=-4, a_{5}=-7, a_{6}=-10, a_{7}=-13$

(iii) $a_{1}=-1, a_{n}=\frac{a_{n-1}}{n}, n \geq 2$

Here, we are given that $n \geq 2$

So, the next five terms of this A.P would be $a_{2}, a_{3}, a_{4}, a_{5}$ and $a_{6}$

Now $a_{1}=-1$ .......(1)

So, to find the $a_{2}$ term we use $n=2$, we get,

$a_{2}=\frac{a_{2-1}}{2}$

$a_{2}=\frac{a_{1}}{2}$

$a_{2}=\frac{-1}{2}($ Using 1)

$a_{2}=\frac{-1}{2} \ldots \ldots$(2)

For $a_{3}$, using $n=3$, we get,

$a_{3}=\frac{a_{3-1}}{3}$

$a_{3}=\frac{a_{2}}{3}$

$a_{3}=\frac{\frac{-1}{2}}{3}$ (Using 2)

$a_{3}=\frac{-1}{6} \ldots \ldots$(3)

For $a_{4}$, using $n=4$, we get,

$a_{4}=\frac{a_{4-1}}{4}$

$a_{4}=\frac{a_{3}}{4}$

$a_{4}=\frac{\frac{-1}{6}}{4}$ (Using 3)

$a_{+}=\frac{-1}{24} \ldots$

For $a_{5}$, using $n=5$, we get,

$a_{5}=\frac{a_{5-1}}{5}$

$a_{5}=\frac{a_{4}}{5}$

$a_{5}=\frac{\frac{-1}{24}}{5}(U \operatorname{sing} 4)$

$a_{5}=\frac{-1}{120} \ldots \ldots$(5)

For $a_{6}$, using $n=6$, we get,

$a_{6}=\frac{a_{6-1}}{6}$

$a_{6}=\frac{a_{5}}{6}$

$a_{6}=\frac{\frac{-1}{120}}{6}(U \operatorname{sing} 5)$

$a_{6}=\frac{-1}{720}$

Therefore, the next five terms, of the given A.P are

$a_{2}=\frac{-1}{2}, a_{3}=\frac{-1}{6}, a_{4}=\frac{-1}{24}, a_{5}=\frac{-1}{120}, a_{6}=\frac{-1}{720}$

(iv) $a_{1}=4, a_{n}=4 a_{n-1}+3, n>1$

Here, we are given that n > 1.

So, the next five terms of this A.P would beand

Now  …… (1)

So, to find theterm we use, we get,

$a_{2}=4 a_{2-1}+3$

$a_{2}=4 a_{1}+3$

$a_{2}=4(4)+3$ (Using 1)

$a_{2}=19 \ldots \ldots(2)$

For $a_{3}$, using $n=3$, we get,

$a_{3}=4 a_{3-1}+3$

$a_{3}=4 a_{2}+3$ (Using 2)

$a_{3}=4(19)+3$

$a_{3}=79 \ldots \ldots(3)$

For $a_{4}$, using $n=4$, we get,

$a_{4}=4 a_{4-1}+3$

$a_{4}=4 a_{3}+3$

$a_{4}=4(79)+3($ Using 3$)$

$a_{4}=319$......(4)

For $a_{5}$, using $n=5$, we get,

$a_{5}=4 a_{5-1}+3$

$a_{5}=4 a_{4}+3$

$a_{5}=4(319)+3($ Using 4$)$

$a_{5}=1279 \ldots \ldots(5)$

For $a_{6}$, using $n=6$, we get,

$a_{6}=4 a_{6-1}+3$

$a_{6}=4 a_{5}+3$

$a_{6}=4(1279)+3($ Using 5$)$

$a_{6}=5119$

Therefore, the next five terms, of the given A.P are

$a_{2}=19, a_{3}=79, a_{4}=319, a_{5}=1279, a_{6}=5119$