Question

The tenth term of an A.P, is -37 and the sum of its first six terms is -27. Find the sum of its first eight terms.

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

• Tenth term of an A.P. is -37.
• Sum of it's first six terms is -27.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The sum of it's first 8th term.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

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

$\longrightarrow\rm{a+(10-1)d=-37}\\\\\longrightarrow\rm{a+9d=-37}\\\\\longrightarrow\rm{a=-37-9d...................(1)}$

&

We know that formula of the sum of an A.P;

$\boxed{\bf{S_n=\frac{n}{2} \bigg[2a+(n-1)d\bigg]}}}}$

$\longrightarrow\rm{\cancel{\dfrac{6}{2}} \bigg[2a+(6-1)d\bigg]=-27}\\\\\\\longrightarrow\rm{3[2a+5d]=-27}\\\\\\\longrightarrow\rm{2a+5d=-27/3}\\\\\\\longrightarrow\rm{2a+5d=-9}\\\\\\\longrightarrow\rm{2(-37-9d)+5d=-9}\\\\\\\longrightarrow\rm{-74-18d+5d=-9}\\\\\\\longrightarrow\rm{-74-13d=-9}\\\\\\\longrightarrow\rm{-13d=-9+74}\\\\\\\longrightarrow\rm{-13d=65}\\\\\\\longrightarrow\rm{d=\cancel{\dfrac{65}{-13}}}\\\\\\\longrightarrow\rm{\pink{d=-5}}$

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

$\longrightarrow\rm{a=-37-9(-5)}\\\\\longrightarrow\rm{a=-37+45}\\\\\longrightarrow\rm{\pink{a=8}}$

Thus;

$\longrightarrow\rm{S_8=\cancel{\dfrac{8}{2}} \bigg[2(8)+(8-1)(-5)\bigg]}\\\\\\\longrightarrow\rm{S_8=4[16+7\times (-5)]}\\\\\\\longrightarrow\rm{S_8=4[16+(-35)]}\\\\\\\longrightarrow\rm{S_8=4[16-35}]}\\\\\\\longrightarrow\rm{S_8=4[-19]}\\\\\\\longrightarrow\rm{\pink{S_8=-76}}$