Progressions MCQ

7^th term = 6;

21^st term = -22;

That means, 14 times common difference or -28 is added to 6 to get -22;

Thus, d = -2;

7^st term = 6 = a+6d;

Or, a+(6*-2) = 6;

Or, a = 18;

26^st term = a+25d = 18-25*2 = -32.

\begin{align} & {\text{We will it through option checking method}}: \cr & \left( {\frac{5}{9}} \right) \times \left( {{{10}^n} - 1} \right) \cr & {\text{We}}\, {\text{put}}\, n = 1, \cr & \left( {\frac{5}{9}} \right) \times \left( {{{10}^1} - 1} \right) = 5 \cr & n = 2\left( {\frac{5}{9}} \right) \times \left( {{{10}^2} - 1} \right) = 55 \cr & n = 3\left( {\frac{5}{9}} \right) \times \left( {{{10}^3} - 1} \right) = 555 \cr & {\text{It means Option C is satisfying the sequence so the}} \cr & {n^{th}}\, {\text{term}}\, {\text{would}}\, {\text{be}}\, \left( {\frac{5}{9}} \right) \times \left( {{{10}^n} - 1} \right) \cr\end{align}

\begin{align} & {\text{Number}}\, {\text{of}}\, {\text{terms}}, \cr & = \left\{ {\frac{{\left( {{1^{st}}\, t{\text{erm - last}}\, {\text{term}}} \right)}}{{{\text{common}}\, {\text{difference}}}}} \right\} + 1 \cr & = \left( {140 - \frac{{20}}{5}} \right) + 1 \cr & = \left( {\frac{{120}}{5}} \right) + 1 \cr & = 24 + 1 \cr & = 25 \cr\end{align}

\begin{align} & {\text{Number}}\, {\text{of}}\, {\text{terms}} = n\, \left( {let} \right) \cr & {\text{First}}\, {\text{term}}\, \left( a \right) = 22 \cr & {\text{Last}}\, {\text{term}}\left( l \right) = - 11 \cr & {\text{Sum}} = 66 \cr & {\text{Sum}}\, {\text{of}}\, {\text{an}}\, {\text{AP}}\, {\text{is}}\, {\text{given}}\, {\text{by}}: \cr & = {\text{Number}}\, {\text{of}}\, {\text{terms}} \times \left\{ {\frac{{\left( {{\text{First}}\, {\text{term}} + {\text{Last}}\, {\text{term}}} \right)}}{2}} \right\} \cr & 66 = n \times \left\{ {\frac{{\left( {a + l} \right)}}{2}} \right\} \cr & 66 = n \times \frac{{\left( {22 - 11} \right)}}{2} \cr & 66 = n \times \left( {\frac{{11}}{2}} \right) \cr & n = \frac{{\left( {66 \times 2} \right)}}{{11}} \cr & n = 12 \cr & {\text{No}}{\text{.}}\, {\text{of}}\, {\text{terms}} = 12 \cr\end{align}

\begin{align} & {5^{th}}\, {\text{term}}\, {\text{of}}\, {\text{GP}} \cr & = a{r^{5 - 1}} \cr & = 16 \times {r^4} \cr & = 81 \cr & {\text{Or}}, \, r = {\left( {\frac{{81}}{{16}}} \right)^{\frac{1}{4}}} = \frac{3}{2} \cr & {4^{th}}\, {\text{term}}\, {\text{of}}\, {\text{GP}} \cr & = a{r^{4 - 1}} \cr & = 16 \times {\left( {\frac{3}{2}} \right)^3} \cr & = 54 \cr\end{align}