Question

The sum of the third and tenth term of an AP is 12. Given that the sum of the 6th terms is 3 less than one third of the eight term. Find the first term and the common difference.

Answer 1

Let a be first term and d- difference of arithmetic  progression

Then $a_3=a+2\cdot d$ - the third term

$a_{10}=a+9\cdot d$ - the tenth term

Therefore from the first condition we have an eqution^

$a+2d+a+9d=2a+11d=12$

The sum of 6th terms is $a_1+a_2+a_3+a_4+a_5+a_6=\frac{a_1+a_6}{2}\cdot 6=$

$=(a_1+a_6)\cdot 3=(2a+5d)\cdot 3=6a+15d$

So second condition can be written as $6a+15d=\frac{a_8}{3}-3=\frac{a+7d}{3}-3$

or as $17a+38d= -9$

Now we have the system:

$\begin{cases} 2a+11d=12\\ 17a+38d= -9 \end{cases}$

From the first eqution we have $a=6-\frac{11}{2}\cdot d$

After we insert a in second eqution

$17a+38d=17\left( 6-\frac{11}{2}\cdot d \right)+38\cdot d=-9\\ 102-\frac{187}{2}\cdot d+38\cdot d=-9;\\ 204-187\cdot d+76\cdot d=-18;\\ 222=111\cdot d; d=\frac{222}{111}=2; a=6-\frac{11}{2}\cdot 2=6-11=-5;\\$

So answer is:

a=-5- the first term of AP;

d=2- difference of AP;