Question

Show that $a_1$,$a_2$,.....,$a_n$ form an AP where an is defined as below:
(i) $a_n$=3+4n
(ii) $a_n$=9−5n
Also find the sum of the first 15 terms in each case.

Answer 1

(i) We need to show that $a_1$,$a_2$...$a_n$ form an AP where $a_n$=3+4n

Let’s calculate values of $a_1$,$a_2$,$a_3$... using $a_n$=3+4n.

$a_1$=3+4(1) =3+4=7

$a_2$=3+4(2) =3+8=11

$a_3$=3+4(3) =3+12=15

$a_4$=3+4(4) =3+16=19

So, the sequence is of the form 7,11,15,19...

Let’s check difference between consecutive terms of this sequence.

11- 7 = 4

15- 11 = 4

19 - 15 = 4

Therefore, the difference between consecutive terms is constant which means terms $a_1$ ,$a_2$.....,$a_n$ form an AP.

We have sequence 7,11,15,19...

First term = a =7

Common difference = d = 4

Applying formula, $S_n$=$\frac{n}{2}$(2a+(n−1)d) to find sum of n terms of AP , we get

$S_{15}$=$\frac{15}{2}$(14+(15−1)4)=$\frac{15}{2}$(14+56)=15$\times$35=525

Therefore, sum of first 15 terms of AP is equal to 525.

(ii) We need to show that $a_1$, $a_2$... $a_n$ form an AP where $a_n$=9−5n

Let’s calculate values of $a_1$,$a_2$,$a_3$... using $a_n$=9−5n.

$a_1$=9−5 (1) =9−5 =4

$a_2$=9−5 (2) = 9−10= −1

$a_3$=9−5 (3) =9−15= −6

$a_4$=9−5 (4) = 9−20= −11

So, the sequence is of the form 4, −1, −6, −11...

Let’s check difference between consecutive terms of this sequence.

-1-(4) = -5

-6-(-1) = -6 + 1 = -5

-11-(-6) = -11 + 6 = -5

Therefore, the difference between consecutive terms is constant which means terms $a_1$,$a_2$,.....,$a_n$ form an AP.

We have sequence 4, −1, −6, −11...

First term = a =4

Commom difference = d = -5

Applying formula, $S_n$=$\frac{n}{2}$(2a+ (n−1) d) to find sum of n terms of AP , we get

$S_{15}$=$\frac{15}{2}$(8+ (15−1) (−5)) =$\frac{15}{2}$(8−70) =15 $\times$ (−31) =−465

Therefore, sum of first 15 terms of AP is equal to -465.