# Show that $a_{1}, a_{2}, \ldots a_{n}, \ldots$ form an $A P$ where $a_{n}$ is defined as below :

Question.

Show that $a_{1}, a_{2}, \ldots a_{n}, \ldots$ form an $A P$ where $a_{n}$ is defined as below :

(i) $a_{n}=3+4 n$

(ii) $a_{n}=9-5 n$

Also find the sum of the first 15 terms in each case.

Solution: (i) $a_{n}=3+4 n$

Putting $\mathrm{n}=1,2,3,4, \ldots$ in $(1)$, we get

$a_{1}=3+4=7, a_{2}=3+8=11$

$a_{3}=3+12=15, a_{4}=3+16=19, \ldots$

Thus, the sequence (list of numbers) is

7, 11, 15, 19, .....

Here, $\quad a_{2}-a_{1}=11-7=4$

$a_{3}-a_{2}=15-11=4$

$a_{4}-a_{3}=19-15=4$

Therefore, the sequence forms an AP in which

a = 7 and d = 4.

$S_{15}=\frac{15}{2}\{2 a+14 d\}=\frac{15}{2}\{2 \times 7+14 \times 4\}$

$=\frac{15}{2} \times 70=15 \times 35=525$

(ii) $a_{n}=9-5 n$

$a_{1}=9-5 \times 1=9-5=4$

$a_{2}=9-5 \times 2=9-10=-1$

$a_{3}=9-5 \times 3=9-15=-6$

$a_{4}=9-5 \times 4=9-20=-11$

It can be observed that

$a_{2}-a_{1}=-1-4=-5$

$a_{3}-a_{2}=-6-(-1)=-5$

$a_{4}-a_{3}=-11-(-6)=-5$

Therefore, this is an A.P. with common difference as $-5$ and first term as 4 .

$S_{15}=\frac{15}{2}[2 a+(n-1) d]$

$=\frac{15}{2}[8+14(-5)]$

$=\frac{15}{2}(8-70)$

$=\frac{15}{2}(-62)=15(-31)$

$=-465$