Mathematics - Sequence & Series Question with Solution | TestHub

Let$T_{\left(1, m\right)}:$$m^{\mathrm{t}\mathrm{h}}$term of$1^{\mathrm{s}\mathrm{t}}$series$A\left(1;3\right)$
$T_{\left(2, n\right)}:$${\mathrm{n}}^{\mathrm{t}\mathrm{h}}$term of$2^{\mathrm{n}\mathrm{d}}$series$A\left(2;5\right)$
$T_{\left(3, r\right)}:$$r^{\mathrm{t}\mathrm{h}}$term of$3^{\mathrm{r}\mathrm{d}}$series$A(3;7)$
For common terms
$T_{(1, m)}=T_{(2, n)}=T_{(3, r)}$
$1+3\left(m-1\right)=2+5\left(n-1\right)=3+7\left(r-1\right)$
$3m-2=5n-3=7r-4$
$\Rightarrow m=\frac{5n-1}{3}\mathrm{a}\mathrm{n}\mathrm{d}r=\frac{5n+1}{7}$
For$‘m’$to be a natural number$n=\mathrm{2,5},11,\dots$
And for$‘r’$to be a natural number$n=\mathrm{4,11},\dots$
When$n=11,m=18$and$r=8$
$\Rightarrow$For$1^{\mathrm{s}\mathrm{t}}$common term of three series$A(a,d)$
$a=1^{\mathrm{s}\mathrm{t}}$term of$A\left(\mathrm{1,3}\right)$
$a=1+\left(181\right)3=52$
$\Rightarrow$Common difference of$A\left(a,d\right)=\mathrm{L}\mathrm{C}\mathrm{M}$of common difference of$\left(A\left(1;3\right),A\left(2;5\right),A\left(3;7\right)\right)$
$d=\mathrm{L}\mathrm{C}\mathrm{M} (3, 5, 7)$
$d=105$
Hence,$a+d=52+105=157$