Question

an ap 4 th term is 11 and 8 term is 5 more than twice the 4 term find the terms of ap​

Answer 1

Solution :

$\bigstar$ Firstly, we know that formula of the A.P;

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

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

A/q

$\longrightarrow\sf{a_4=11}\\\\\longrightarrow\sf{a+(n-1)d=11}\\\\\longrightarrow\sf{a+(4-1)d=11}\\\\\longrightarrow\sf{a+3d=11......................(1)}$

&

$\pi \longrightarrow\sf{a_8=5+2(4^{th}\:term)}\\\\\longrightarrow\sf{a+(8-1)d=5+2(a+3d)}\\\\\longrightarrow\sf{a+7d=5+2a+6d}\\\\\longrightarrow\sf{a-2a+7d-6d=5}\\\\\longrightarrow\sf-a+d=5}\\\\\longrightarrow\sf{d=5+a.....................(2)}$

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

$\longrightarrow\sf{a+3(5+a)=11}\\\\\longrightarrow\sf{a+15+3a=11}\\\\\longrightarrow\sf{4a+15=11}\\\\\longrightarrow\sf{4a=11-15}\\\\\longrightarrow\sf{4a=-4}\\\\\longrightarrow\sf{a=\cancel{-4/4}}\\\\\longrightarrow\bf{a=-1}$

Putting the value of [a] in equation (2),we get;

$\longrightarrow\sf{d=5+(-1)}\\\\\longrightarrow\sf{d=5-1}\\\\\longrightarrow\bf{d=4}$

Now;

$\boxed{\boldsymbol{Arithmetic\:Progression\::}}}$

$\bullet\:\sf{a=\boxed{\bf{-1}}}\\\\\bullet\sf{a+d=-1+4=\boxed{\bf{3}}}\\\\\bullet\sf{a+2d=-1+2(4)=-1+8=\boxed{\bf{7}}}\\\\\bullet\sf{a+3d=-1+3(4)=-1+12=\boxed{\bf{11}}}\\\\\bullet\sf{a+4d=-1+4(4)=-1+16=\boxed{\bf{15}}}$