Question

There are ten numbers in A.P.. If the sum of first three terms is $321$ and the sum of last three numbers is $405$, then the sum of all ten numbers is

• A. $1180$
• B. $1210$
• C. $1255$
• D. $1270$

Answer 1

The correct option is B $1210$

Let the first term be $a$ and the common difference be $d.$
According to given condition, we get
$T_1+T_2+T_3=321T_8+T_9+T_{10}=405$
Now,
$T_1+T_2+T_3=321\Rightarrow a+(a+d)+(a+2d)=321\Rightarrow a+d=107\cdots \left(1\right)$

Also,
$T_8+T_9+T_{10}=405\Rightarrow (a+7d)+(a+8d)+(a+9d)=405\Rightarrow a+8d=135\cdots \left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$, we get
$d=4,a=103$
Now, sum of all $10$ terms,
$S_{10}=\frac{10}{2}[2a+9d]\Rightarrow S_{10}=5[206+36]\therefore S_{10}=1210$