Progressions

1.How many terms are there in 20, 25, 30......... 140

1. 22
2. 25
3. 23
4. 24

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Number of terms, \\ = {\frac{(1^{s t} t erm - last term)}{common difference}} + 1 \\ = (140 - \frac{20}{5}) + 1 \\ = (\frac{120}{5}) + 1 \\ = 24 + 1 \\ = 25 \end{aligned}

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

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2.Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.

1. 5
2. 6
3. 4
4. 3

Answer:Option 1

Explanation:

Solution:

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.

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3.Find the 15th term of the sequence 20, 15, 10....

1. -45
2. -55
3. -50
4. 0

Answer:Option 1

Explanation:

Solution:

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

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4.How many terms are there in the GP 5, 20, 80, 320........... 20480?

1. 5
2. 6
3. 8
4. 7

Answer:Option 1

Explanation:

Solution:

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.

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5.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?

1. 2²⁰
2. 2²⁰-1
3. 2^{19^-1}
4. 2¹⁹

Answer:Option 1

Explanation:

Solution:

\begin{aligned} 1^{s t} term = 1; \\ Common ration = 2 \\ Sum (S_{n}) = a \times \frac{(r^{n} - 1)}{(r - 1)} \\ = 1 \times \frac{(2^{20} - 1)}{(2 - 1)} \\ = 2^{20} - 1 \end{aligned}

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

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6.If the fifth term of a GP is 81 and first term is 16, what will be the 4^th term of the GP?

1. 36
2. 18
3. 54
4. 24

Answer:Option 1

Explanation:

Solution:

\begin{aligned} 5^{t h} term of GP \\ = a r^{5 - 1} \\ = 16 \times r^{4} \\ = 81 \\ Or, r = {(\frac{81}{16})}^{\frac{1}{4}} = \frac{3}{2} \\ 4^{t h} term of GP \\ = a r^{4 - 1} \\ = 16 \times {(\frac{3}{2})}^{3} \\ = 54 \end{aligned}

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

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7.The 7^th and 21^st terms of an AP are 6 and -22 respectively. Find the 26^th term.

1. -34
2. -32
3. -12
4. -10

Answer:Option 1

Explanation:

Solution:

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.

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8.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 :

1. 3¹⁰ /2
2. 3¹⁰ - 2¹⁰
3. 243 *(3⁵ -1)
4. 3¹⁰ -2⁵

Answer:Option 1

Explanation:

Solution:

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.

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9.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.

1. 32 Cm²
2. 16 Cm²
3. 20 Cm²
4. 64 Cm²

Answer:Option 1

Explanation:

Solution:

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

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

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10.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:

1. 10
2. 12
3. 9
4. 8

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Number of terms = n (l e t) \\ First term (a) = 22 \\ Last term (l) = - 11 \\ Sum = 66 \\ Sum of an AP is given by : \\ = Number of terms \times {\frac{(First term + Last term)}{2}} \\ 66 = n \times {\frac{(a + l)}{2}} \\ 66 = n \times \frac{(22 - 11)}{2} \\ 66 = n \times (\frac{11}{2}) \\ n = \frac{(66 \times 2)}{11} \\ n = 12 \\ No . of terms = 12 \end{aligned}

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

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