Progressions - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{terms}}, \cr & = {\frac{{ {{1^{st}}\,t{\text{erm - last}}\,{\text{term}}} }}{{{\text{common}}\,{\text{difference}}}}} + 1 \cr & = { \frac{{140 -20}}{5}} + 1 \cr & = {\frac{{120}}{5}} + 1 \cr & = 24 + 1 \cr & = 25 \cr} $$

1^st Method:
8^th term = a + 7d = 39 ........... (i)
12^th 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
2^nd Method (Thought Process):
8^th term = 39
And, 12^th term = 59
Here, we see that 20 is added to 8^th term 39 to get 12^th term 59 i.e. 4 times the common difference is added to 39
So, CD = $$\frac{{20}}{4}$$  = 5
Hence, 7 times CD is added to 1^st term to get 39. That means 4 is the 1^st term of the AP

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

$$\eqalign{ & {1^{st}}{\kern 1pt} {\text{Method}}: \cr & {1^{st}}{\kern 1pt} {\text{term}} = 5; \cr & {3^{rd}}{\kern 1pt} {\text{term}} = 15; \cr & {\text{Then}},{\kern 1pt} \,d = 5; \cr & {16^{th}}{\kern 1pt} {\text{term}} = a + 15d \cr & = 5 + 15 \times 5 = 80 \cr & {\text{Sum}} = {n \times \frac{{\left( {a + l} \right)}}{2}} \cr} $$
$$ = {{\text{no}}{\text{.}}{\kern 1pt} {\text{of}}{\kern 1pt} {\text{terms}} \times \frac{{ {{\text{first}}{\kern 1pt} {\text{term + last}}{\kern 1pt} {\text{term}}} }}{2}} $$
$$\eqalign{ & = {16 \times \frac{{\left( {5 + 80} \right)}}{2}} \cr & = 16 \times \frac{{85}}{2} \cr & = 8 \times 85 \cr & = 680 \cr} $$

2^nd Method(Thought Process):
Sum = number of terms × average of that AP
$$\eqalign{ & {\text{Sum}} = 16 \times {\frac{{\left( {5 + 80} \right)}}{2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 16 \times \frac{{85}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 8 \times 85 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 680 \cr} $$

Common ratio, r = $$\frac{{20}}{5}$$ = 4
Last term or n^th term of GP = ar^{n - 1}
20480 = 5 × (4^{n - 1})
Or, 4^{n - 1} = $$\frac{{20480}}{5}$$  = 4⁶
So, comparing the power,
Thus, n - 1 = 6
Or, n = 7
Number of terms = 7

$$\eqalign{ & {1^{st}}\,{\text{term}} = 1; \cr & {\text{Common}}\,{\text{ration}} = 2 \cr & {\text{Sum}}\left( {{S_n}} \right) = a \times \frac{{ {{r^n} - 1} }}{{ {r - 1} }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \times \frac{{ {{2^{20}} - 1} }}{{ {2 - 1} }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {2^{20}} - 1 \cr} $$

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

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

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:
$$\eqalign{ & \frac{a}{{ {1 - r} }} \cr & {\text{Now}}, \cr & { {\frac{{120}}{{ {\frac{1}{5}} }}} + {\frac{{96}}{{ {\frac{1}{5}} }}} } \cr & = 1080\,m \cr} $$

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.