Question

The sum of 6 terms which form an A.P. is 345. The difference between the first and last term is 55. Find the terms.

Answer 1

Let the terms be $a,a+d,a+2d,a+3d,a+4d,a+5d$.

It is given that the difference between the first and last term is $55$, therefore,

$a+5d-a=55\Rightarrow 5d=55\Rightarrow d=\frac{55}{5}=11......\left(1\right)$

It is also given that the sum of all terms is $345$, therefore using equation 1 we have,

$a+a+d+a+2d+a+3d+a+4d+a+5d=345\Rightarrow 6a+15d=345\Rightarrow 6a+(15\times 11)=345\Rightarrow 6a+165=345\Rightarrow 6a=345-165\Rightarrow 6a=180\Rightarrow a=\frac{180}{6}=30$

With $a=30$ and $d=11$, the $6$ terms are as follows:

$a=30$

$a+d=30+11=41$

$a+2d=30+(2\times 11)=30+22=52$

$a+3d=30+(3\times 11)=30+33=63$

$a+4d=30+(4\times 11)=30+44=74$

$a+5d=30+(5\times 11)=30+55=85$

Hence, the terms are $30,41,52,63,74,85$.