Question

Determine the AP whose third term is 5 and 7 term in 9

Answer 1

Solution :

$\bigstar$ 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 number.

$\underline{\boldsymbol{According\:to\:the\:question\::}}}$

$\longrightarrow\sf{a_3=5}\\\\\longrightarrow\sf{5=a+(3-1)d}\\\\\longrightarrow\sf{5=a+2d..........................(1)}$

&

$\longrightarrow\sf{a_7=9}\\\\\longrightarrow\sf{9=a+(7-1)d}\\\\\longrightarrow\sf{9=a+6d..........................(2)}$

$\underline{\boldsymbol{Using\:by\:Substitution\:method\::}}}$

From equation (2),we get;

$\longrightarrow\sf{9=a+6d}\\\\\longrightarrow\sf{a=9-6d...........................(3)}$

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

$\longrightarrow\sf{5=9-6d+2d}\\\\\longrightarrow\sf{5-9=-4d}\\\\\longrightarrow\sf{-4=-4d}\\\\\longrightarrow\sf{d=\cancel{-4/-4}}\\\\\longrightarrow\bf{d=1}$

∴ Putting the value of d in equation (3),we get;

$\longrightarrow\sf{a=9-6(1)}\\\\\longrightarrow\sf{a=9-6}\\\\\longrightarrow\bf{a=3}$

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

$\bullet\:\sf{a=\boxed{\bf{3}}}\\\\\bullet\sf{a+d=3+1=\boxed{\bf{4}}}\\\\\bullet\sf{a+2d=3+2(1)=3+2=\boxed{\bf{5}}}\\\\\bullet\sf{a+2d=3+3(1)=3+3=\boxed{\bf{6}}}$