In an AP, Tm=1/25, T25=1/20 and 20 ∑ Tr r ∈ to (r=1, 25) = 13, then 5m ∑ Tr r ∈ to (r=m, 2m) equals

Question

In an AP, \(T_m=\frac{1}{25}, T_{25}=\frac{1}{20}\) and \(20 \sum_{r=1}^{\mathrm{25}} T_r=13,\) then \(5 m \sum_{r=m}^{2 m} T_r \text { equals }\)

Answer 1

Answer is "126"  

\(T_m=a+(m-1) d \cdot \frac{1}{25}\)   ...(1)

\(T_{\mathrm{25}}=a+24 d=\frac{1}{20}\)   

\(20 \times \frac{2 r}{2}\left[a+\frac{1}{20}\right]=13 \Rightarrow a=\frac{1}{20 \times 25}\)  

\(20 \sum_{r=1}^{25} T=20 \times \frac{25}{2}[2 a+24 d]=13\)

\(d=\frac{1}{20 \times 25}\)  

Substitute a and din (i)  

\(\Rightarrow m=20\)  

Now \(5 m \sum_{r=m}^{2 m} T_r=5 \times 20\left[\sum_{r=m 0}^{40} T_r\right]\)