Question

find the AP of 18 15 12 – – – - 45 and also find the the sum of all its item​

Answer 1

Correct Question :-

Find the number of terms of the AP of 18 , 15 , 12 --- - - ( - 45 ) and also find the sum of all its term.

Solution :-

Given AP = 18 , 15 , 12 ... ... ... -45

Therefore ,

First term a = 18

Common difference d = - 3

Last term an = 45

Now,

We have to find the number of terms

Therefore,

By using an formula, we get :-

an = a + ( n - 1 )d

Subsitute the required values,

- 45 = 18 + ( n - 1 )-3

-45 = 18 + ( n - 1 )-3

-45 - 18 = (n - 1 )-3

-63 = ( n - 1 )-3

-63/-3 = n - 1

21 = n - 1

21 + 1 = n

22 = n

n = 22

Thus, The total number of terms are 22

Now,

We have to find the sum of terms

Therefore,

By using Sn formula we get :-

Sn = n/2 ( 2a + ( n - 1 )d)

Subsitute the required values,

Sn = 22/2 ( 2 * 18 + ( 22 - 1 )-3 )

Sn = 11 ( 36 + 21 * ( - 3 ))

Sn = 11 ( 36 - 63 )

Sn = 11 ( -27)

Sn = -297

Hence, The sum of the terms of an given AP is -297 $.$

Answer 2

Step-by-step explanation:

$\sf Correct\:Question\begin{cases}\rm\leadsto Find\:the\:number\:of\:terms\:of\:the\:A.P, \\ \rm \:\;\;\;18,15,12\dots -45.And\:also\:Find\:the\:sum\:of\:them.\end{cases}$

SOLUTION:-

• 18,15,12.. .. -45 are in AP

Here

• a=18
• d=18-15=3
• $\bf t_n=-45$

ATQ,

$\\ \tt{:}\longrightarrow t_n=-45$

$\\ \tt{:}\longrightarrow a+(n-1)d=-45$

$\\ \tt{:}\longrightarrow 18+(n-1)-3=-45$

$\\ \tt{:}\longrightarrow 18-3n+3=-45$

$\\ \tt{:}\longrightarrow -3n+21=-45$

$\\ \tt{:}\longrightarrow -3n=-45-21=-66$

$\\ \tt{:}\longrightarrow n=\dfrac{-66}{-3}$

$\\ \tt{:}\longrightarrow n=22$

Now ,

$\boxed{\sf S_n=\dfrac{n}{2}\left\{2a+(n-1)d\right\}}$

$\\ \tt{:}\longrightarrow S_{22}=\dfrac{22}{2}\left\{2(18)+(22-1)-3\right\}$

$\\ \tt{:}\longrightarrow S_{22}=11\left\{36+21(-3)\right\}$

$\\ \tt{:}\longrightarrow S_{22}=11(36-63)$

$\\ \tt{:}\longrightarrow S_{22}=11(-27)$

$\\ \boxed{\bf{:}\longrightarrow S_{22}=-297}$