Question

An AP starts with a positive fraction and every alternate term is an integer. If the sum of the first 11 terms is 33, then find the 4th term

Answer 1

Dear student
$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that} \\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right] \\ \mathrm{So}, \\ {\mathrm{S}}_{11}=\frac{11}{2}\left[2\mathrm{a}+\left(11-1\right)\mathrm{d}\right] \\ 33=\frac{11}{2}\left[2\mathrm{a}+10\mathrm{d}\right] \\ \frac{66}{11}=2\mathrm{a}+10\mathrm{d} \\ 2\mathrm{a}+10\mathrm{d}=6 \\ \mathrm{a}+5\mathrm{d}=3...\left(1\right) \\ \mathrm{Since}\mathrm{every}\mathrm{alternate}\mathrm{term}\mathrm{is}\mathrm{an}\mathrm{integer}\mathrm{and}\mathrm{given}\mathrm{sum}\mathrm{is}\mathrm{positive} \\ \mathrm{a}+3\mathrm{d}=2...\left(2\right) \\ \mathrm{Subtracting}\left(2\right)\mathrm{and}\left(1\right) \\ 2\mathrm{d}=1 \\ \mathrm{d}=\frac{1}{2} \\ \mathrm{a}=2-\frac{3}{2} \\ \mathrm{a}=\frac{4-3}{2} \\ \mathrm{a}=\frac{1}{2} \\ \mathrm{So},{\mathrm{a}}_4=\mathrm{a}+3\mathrm{d} \\ =\frac{1}{2}+3\times \left(\frac{1}{2}\right) \\ =\frac{1}{2}+\frac{3}{2} \\ =\frac{4}{2} \\ =2 \end{gathered}$$
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