If sum of the sequence 1,6,11,16,....,x …

Given series is

{tex} 1+6+11+16+\ldots . .+x=148 Here, a=1, d=6-1=11-6=16-11=5 and \mathrm{S}_{\mathrm{n}}=148 \begin{aligned} & \because \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ & \therefore 148=\frac{\mathrm{n}}{2}[2 \times 1+(\mathrm{n}-1) 5] \\ & \Rightarrow 296=2 \mathrm{n}+5 \mathrm{n}^2-5 \mathrm{n} \\ & \Rightarrow 296=5 \mathrm{n}^2-3 \mathrm{n} \\ & \Rightarrow 5 \mathrm{n}^2-3 \mathrm{n}-296=0 \\ & \Rightarrow(\mathrm{n}-8)(5 \mathrm{n}+37)=0 \\ & \Rightarrow \mathrm{n}=8, \mathrm{n}=\frac{-37}{5} \end{aligned} {/tex}

{tex}\begin{aligned} & \therefore \mathrm{n}=8 \\ & (\because \mathrm{n} \text { cannot be negative }) \\ & \text { Now, } \mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} \\ & \mathrm{x}=1+(8-1) \times 5 \\ & \Rightarrow \mathrm{x}=1+35 \Rightarrow \mathrm{x}=36 . \end{aligned}{/tex}