Question

determine the AP whose fourth term is 18 and the difference of the 9th term from the 15 term is 30​

Answer 1

Answer:

3,8,13,18,23.......

Step-by-step explanation:

a4=18

a+3d=18

a=18-3d------(1)

a15-a9=a+14d-1-8d=30

6d=30

d=5

From(1) a=8-15=3

AP=3,8,13,18,23.....

Answer 2

Solution :

Firstly, we know that formula of an A.P;

$\boxed{\bf{a_n=a+(n-1)d}}}$

• a is the first term.
• d is the common difference.
• n is the term of an A.P.

A/q

$\longrightarrow\sf{a_4=18}\\\\\longrightarrow\sf{a+(4-1)d=18}\\\\\longrightarrow\sf{a+3d=18}\\\\\longrightarrow\sf{a=18-3d ..................(1)}$

&

$\longrightarrow\sf{a_{15} - a_{9} = 30}\\\\\longrightarrow\sf{[a+(15-1)d]-[a+(9-1)d] =30}\\\\\longrightarrow\sf{[a+14d] - [a + 8d] = 30}\\\\\longrightarrow\sf{\cancel{a}+14d \cancel{- a }- 8d = 30}\\\\\longrightarrow\sf{14d-8d = 30}\\\\\longrightarrow\sf{6d = 30}\\\\\longrightarrow\sf{d=\cancel{30/6}}\\\\\longrightarrow\bf{d=5}$

∴ Putting the value of d in equation (1),we get;

$\longrightarrow\sf{a=18-3(5)}\\\\\longrightarrow\sf{a=18-15}\\\\\longrightarrow\bf{a=3}$

Thus;

$\boxed{\bf{Arithmetic\:progression\::}}}$

$\bullet\:\sf{a=\boxed{\bf{3}}}\\\\\bullet\sf{a+d=3+5=\boxed{\bf{8}}}\\\\\bullet\sf{a+2d=3+2(5)=3+10=\boxed{\bf{13}}}\\\\\bullet\sf{a+3d=3+3(5)=3+15=\boxed{\bf{18}}}$