Write first 4 terms in each of the sequences:

Solution:

To Find: First four terms of given series.

(i) Given: $\mathrm{n}^{\text {th }}$ term of series is $(5 \mathrm{n}+2)$

Put $n=1,2,3,4$ in $n^{\text {th }}$ term, we get first (a1), Second (a2), Third (a3) \& Fourth (a4) term

$a_{1}=(5 \times 1+2)=7$

$a_{2}=(5 \times 2+2)=12$

$a_{3}=(5 \times 3+2)=17$

$a_{4}=(5 \times 4+2)=22$

First four terms of given series is 7, 12,17,22

ALTER: When you find or you have first term $\left(a\right.$ or $\left.a_{1}\right)$ and second term $\left(a_{2}\right)$ then find the difference $\left(a_{2}-a_{1}\right)$

Now add this difference in last term to get the next term

For example $a_{1}=7$ and $a_{2}=12$, so difference is $12-5=7$

Now $\mathrm{a}_{3}=12+5=17, \mathrm{a}_{4}=17+5=22$

(This method is only for A.P)

NOTE: When you have nth term in the form of $(a \times n+b)$

Then common difference of this series is equal to a.

This type of series is called A.P (Arithmetic Progression)

(Where $a, b$ are constant, and $n$ is number of terms)

(ii) Given: $n^{\text {th }}$ term of series is $\frac{(2 n-3)}{4}$

Put n=1, 2, 3, 4 in nth term, we get first (a1), Second (a2), Third (a3) & Fourth (a4) term

$a_{1}=\frac{(2 \times 1-3)}{4}=\frac{-1}{4}$

$a_{2}=\frac{(2 \times 2-3)}{4}=\frac{1}{4}$

$a_{3}=\frac{(2 \times 3-3)}{4}=\frac{3}{4}$

$a_{4}=\frac{(2 \times 4-3)}{4}=\frac{5}{4}$

First four terms of given series are $\frac{-1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{5}{4}$

(iii) Given: $n^{\text {th }}$ term of series is $(-1)^{n-1} \times 2^{n+1}$

Put $n=1,2,3,4$ in $n^{\text {th }}$ term, we get first $(a 1)$, Second $(a 2)$, Third $(a 3) \&$ Fourth $(a 4)$ term.

$a_{1}=(-1)^{1-1} \times 2^{1+1}=(-1)^{0} \times 2^{2}=1 \times 4=4$

$a_{2}=(-1)^{2-1} \times 2^{2+1}=(-1)^{1} \times 2^{3}=(-1) \times 8=(-8)$

$a_{3}=(-1)^{3-1} \times 2^{3+1}=(-1)^{2} \times 2^{4}=1 \times 16=16$

$a_{4}=(-1)^{4-1} \times 2^{4+1}=(-1)^{3} \times 2^{5}=(-1) \times 32=(-32)$

First four terms of given series are $4,-8,16,-32$