Question

Solve the equation $(x+1)+(x+4)+(x+7)+....+(x+28)=155$.

Answer 1

Consider $1,4,7,...,28$ are in A.P

First term$=a=1$

Common difference$=a_2-a_1=4-1=3$

$n^{th}$ term$=a_n=28$

$a+(n-1)d=28$

$\Rightarrow 1+(n-1)3=28$

$\Rightarrow 3(n-1)=28-1=27$

$\Rightarrow n-1=\frac{27}{3}=9$

$\therefore n=9+1=10$

Sum of $n$ terms$=S_n=\frac{n}{2}(a+a_n)$

Here $a=1$, $n=10$

$S_{10}=\frac{10}{2}(1+28)=5\times 29=145$ ......$\left(1\right)$

Now, $(x+1)+(x+4)+(x+7)+...+(x+28)=155$

$[(x+x+x+...+10terms)+(1+4+7+...+28)]=155$

$\Rightarrow 10x+145=155$

$\Rightarrow 10x=155-145=20$

$\therefore x=\frac{10}{10}=1$