Question

The ratio of nth term of two A.P.s is $(14n-6):(8n+23)$, then the ratio of their sum of first m terms is

• A. $\frac{4m+4}{7m+24}$
• B. $\frac{7m+1}{4m+24}$
• C. $\frac{7m+1}{4m+27}$
• D. $\frac{28m-20}{16m+15}$

Answer 1

The correct option is C $\frac{7m+1}{4m+27}$

$\frac{T_n}{T_n^{\prime }}=\frac{14n-6}{8n+23}=\frac{a+(n-1)d}{a^{\prime }+(n-1)d^{\prime }}$
$\frac{S_m}{S_m^{\prime }}=\frac{\frac{m}{2}[2a+(m-1\left)d\right]}{\frac{m}{2}[2a^{\prime }+(m-1\left)d^{\prime }\right]}=\frac{2a+(m-1)d}{2a^{\prime }+(m-1)d^{\prime }}=\frac{a+\frac{m-1}{2}d}{a^{\prime }+\frac{m-1}{2}{d}^{\prime }}$
Substituting $2n-2=m-1$ we get $n=\frac{m+1}{2}$
$\frac{S_m}{S_m^{\prime }}=\frac{14\left(\frac{m+1}{2}\right)-6}{8\left(\frac{m+1}{2}\right)+23}=\frac{7m+1}{4m+27}$