If the pth term of an A.P is 4p-1,find the 40th term and the sum of first 40 terms

Dear Student,

Please find below the solution to the asked query :

$$\begin{gathered} \mathrm{Let} \mathrm{a} \mathrm{be} \mathrm{the} \mathrm{first} \mathrm{term} \mathrm{and} \mathrm{d} \mathrm{be} \mathrm{the} \mathrm{common} \mathrm{difference}. \\ \mathrm{Now}, {\mathrm{a}}_{\mathrm{p}} = 4\mathrm{p} - 1 \\ \mathrm{Now}, {\mathrm{a}}_1 = \mathrm{a} = 4 \times 1 - 1 = 4 - 1 = 3 \\ \mathrm{Now}, {\mathrm{a}}_2 = 4 \times 2 - 1 = 8 - 1 = 7 \\ \mathrm{So}, \mathrm{d} = {\mathrm{a}}_2 - {\mathrm{a}}_1 = 7 - 3 = 4 \\ \mathrm{Now}, {\mathrm{a}}_{40} = \mathrm{a}+39\mathrm{d} = 3 + 39 \times 4 = 3 + 156 = 159 \\ \mathrm{Now}, \mathrm{sum} \mathrm{of} \mathrm{first} \mathrm{n} \mathrm{terms} \mathrm{of} \mathrm{an} \mathrm{AP} \mathrm{is} \\ {\mathrm{S}}_{\mathrm{n}} = \frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right] \\ \mathrm{Now}, {\mathrm{S}}_{40} = \frac{40}{2}\left[2\times 3+39\times 4\right] = 20\left(60+156\right) = 20 \times 216 = 4320 \end{gathered}$$

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