Progressions MCQ

Progressions

Question 1 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
The 7th and 21st terms of an AP are 6 and -22 respectively. Find the 26th term.
A
-32
B
-34
C
-10
D
-16
Question 1 Explanation: 
7th term = 6;

21st term = -22;

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

Thus, d = -2;

7st term = 6 = a+6d;

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

Or, a = 18;

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







Question 2 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
Find the nth term of the following sequence :

5 + 55 +555 + . . . . Tn

A
(5/9)*(10n - 1)
B
5n(10n - 1)
C
(5/9)n *10n - 1)
D
5(10n - 1)
Question 2 Explanation: 
\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}
Question 3 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
How many terms are there in 20, 25, 30..... 140
A
23
B
25
C
22
D
24
Question 3 Explanation: 
\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}






Question 4 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
The first term of an Arithmetic Progression is 22 and the last term is -11. If the sum is 66, the number of terms in the sequence are:
A
8
B
9
C
12
D
10
Question 4 Explanation: 
\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}
Question 5 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
If the fifth term of a GP is 81 and first term is 16, what will be the 4th term of the GP?
A
24
B
18
C
27
D
54
Question 5 Explanation: 
\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}






Question 6 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
Find the 15th term of the sequence 20, 15, 10....
A
-55
B
-50
C
-45
D
0
Question 6 Explanation: 
15th term = a+14d = 20+14*(-5) = 20-70 = -50.
Question 7 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.
A
7
B
6
C
4
D
5
Question 7 Explanation: 
1st 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.







Question 8 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
A square is drawn by joining the mid points of the sides of a given square in the same way and this process continues indefinitely. If a side of the first square is 4 cm, determine the sum of the areas all the square.
A
32 Cm2
B
16 Cm2
C
64 Cm2
D
20 Cm2
Question 8 Explanation: 
\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}
Question 9 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
A bacteria gives birth to two new bacteria in each second and the life span of each bacteria is 5 seconds. The process of the reproduction is continuous until the death of the bacteria. initially there is one newly born bacteria at time t = 0, the find the total number of live bacteria just after 10 seconds :
A
310 /2
B
243 *(35 -1)
C
310 - 210
D
310 -25
Question 9 Explanation: 
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.







Question 10 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is
A
640
B
690
C
680
D
765
Question 10 Explanation: 
\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}
Question 11 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
A boy agrees to work at the rate of one rupee on the first day, two rupees on the second day, and four rupees on third day and so on. How much will the boy get if he started working on the 1st of February and finishes on the 20th of February?
A
220
B
219
C
219-1
D
220-1
Question 11 Explanation: 
\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}






Question 12 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
How many terms are there in the GP 5, 20, 80, 320..... 20480?
A
7
B
6
C
8
D
5
Question 12 Explanation: 
Common ratio, r = 20/5 = 4;

Last term or nth term of GP = arn-1.

20480 = 5*(4n-1);

Or, 4n-1 = 20480/5 = 48;

So, comparing the power,

Thus, n-1 = 8;

Or, n = 7;

Number of terms = 7.

Question 13 ->Click on any option to know the correct answers (सही उत्तर जानने के लिए किसी भी Choice पर क्लिक करें)
After striking the floor, a rubber ball rebounds to 4/5th of the height from which it has fallen. Find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 metres.
A
1020 m
B
1080 m
C
960 m
D
1120 m
Question 13 Explanation: 
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}

There are 13 questions to complete.

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