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}

15^th term = a+14d = 20+14*(-5) = 20-70 = -50.

1^st Method:

8th term = a+7d = 39 ..... (i)

12th term = a+11d = 59 ..... (ii)

(i)-(ii);

Or, a+7d-a-11d = 39-59; Or, 4d = 20;

Or, d = 5;

Hence, a+7*5 = 39;

Thus, a = 39-35 = 4.

2nd Method (Thought Process):

8th term = 39;

And, 12th term = 59;

Here, we see that 20 is added to 8th term 39 to get 12th term 59 i.e. 4 times the common difference is added to 39.

So, CD = 20/4 = 5.

Hence, 7 times CD is added to 1st term to get 39. That means 4 is the 1st term of the AP.

\begin{align} & {\text{Side of the first square is }}4{\text{ }}cm. \cr & {\text{side of second square}} \cr & = 2\sqrt 2 \,cm \cr & {\text{Side}}\,{\text{of}}\,{\text{third}}\,{\text{square}} \cr & = 2\,cm \cr & {\text{and}}\,{\text{so}}\,{\text{on}}{\text{.}}\,i.e. \cr & 4,\,2,\,\sqrt 2 ,\,\sqrt 2 ,\,1\,..... \cr & {\text{Thus, area of these square will be}}, \cr & = 16,\,8,\,4,\,2,\,1,\,\frac{1}{2}..... \cr & {\text{Hence, Sum of the area of first, second, third square}}.... \cr & = 16 + 8 + 4 + 2 + 1 + \,..... \cr & = \left[ {\frac{{16}}{{\left\{ {1 - \left( {\frac{1}{2}} \right)} \right\}}}} \right] \cr & = 32\,c{m^2} \cr\end{align}

Total number of bacteria after 10 seconds,

= 3¹⁰ - 3⁵

= 3⁵ *(3⁵ -1)

= 243 *(3⁵ -1)

Since, just after 10 seconds all the bacterias (i.e. 3⁵ ) are dead after living 5 seconds each.

\begin{align} & {1^{st}}\,{\text{Method}}: \cr & {1^{st}}\,{\text{term}} = 5; \cr & {3^{rd}}\,{\text{term}} = 15; \cr & {\text{Then}},\,d = 5; \cr & {16^{th}}\,{\text{term}} = a + 15d \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5 + 15 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 80 \cr & {\text{Sum}} = \left\{ {n \times \frac{{\left( {a + l} \right)}}{2}} \right\} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ {{\text{no}}{\text{.}}\,{\text{of}}\,{\text{terms}} \times \frac{{\left( {{\text{first}}\,{\text{term + last}}\,{\text{term}}} \right)}}{2}} \right\} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ {16 \times \frac{{\left( {5 + 80} \right)}}{2}} \right\} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 680 \cr & \cr & {2^{nd}}\,{\text{Method}}\,({\text{Thought}}\,{\text{Process}}): \cr & {\text{Sum}} = {\text{number}}\,{\text{of}}\,{\text{terms}} \times {\text{average}}\,{\text{of}}\,{\text{that}}\,{\text{AP}} \cr & {\text{Sum}} = 16 \times \left\{ {\frac{{\left( {5 + 80} \right)}}{2}} \right\} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 16 \times 45 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 680 \cr\end{align}

\begin{align} & {1^{st}}\,{\text{term}} = 1; \cr & {\text{Common}}\,{\text{ration}} = 2 \cr & {\text{Sum}}\left( {{S...n}} \right) = a \times \frac{{\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \times \frac{{\left( {{2^{20}} - 1} \right)}}{{\left( {2 - 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {2^{20}} - 1 \cr\end{align}

Common ratio, r = 20/5 = 4;

Last term or n^th term of GP = ar^n-1.

20480 = 5*(4^n-1);

Or, 4^n-1 = 20480/5 = 4⁸;

So, comparing the power,

Thus, n-1 = 8;

Or, n = 7;

Number of terms = 7.

The first drop is 120 metres. After this the ball will rise by 96 metres and fall by 96 metres. This process will continue in the form of infinite GP with common ratio 0.8 and first term 96. The required answer is given by the formula:

\begin{align} & \frac{a}{{\left( {1 - r} \right)}} \cr & {\text{Now}}, \cr & \left[ {\left\{ {\frac{{120}}{{\left( {\frac{1}{5}} \right)}}} \right\} + \left\{ {\frac{{96}}{{\left( {\frac{1}{5}} \right)}}} \right\}} \right] \cr & = 1080\,m \cr\end{align}