Progressions — Part 1 MCQs 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 = $$\frac{{20480}}{5}$$ = 4 6 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, = 3 10 - 3 5 = 3 5 × (3 5 -1) = 243 × (3 5 -1) Since, just after 10 seconds all the bacterias (i.e. 3 5 ) are dead after living 5 seconds each.

Side of the first square is 4 cm. side of second square = $$2\sqrt 2 {\kern 1pt} $$ cm. Side of third square = 2 cm. 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 $$\eqalign{ & = 16 + 8 + 4 + 2 + 1 + {\kern 1pt} \,...... \cr & = {\frac{{16}}{{ {1 - {\frac{1}{2}} } }}} \cr & = 32\,{\kern 1pt} c{m^2} \cr} $$

Number of terms = n (let) First term (a) = 22 Last term (l) = - 11 Sum = 66 Sum of an AP is given by: $$ = {\text{Number}}\,{\text{of terms}}\,\, \times $$ $$ {\frac{{ {{\text{First}}\,{\text{term}} + {\text{Last}}\,{\text{term}}} }}{2}} $$ $$\eqalign{ & 66 = {\text{n}} \times {\frac{{ {{\text{a}} + {\text{l}}} }}{2}} \cr & 66 = {\text{n}} \times \frac{{ {22 - 11} }}{2} \cr & 66 = {\text{n}} \times {\frac{{11}}{2}} \cr & {\text{n}} = \frac{{ {66 \times 2} }}{{11}} \cr & {\text{n}} = 12 \cr & {\text{No}}{\text{.}}\,{\kern 1pt} {\text{of}}\,{\text{terms}} = 12 \cr} $$

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

$$\eqalign{ & {T_2} = a + d = 17\,.......\,\left( 1 \right) \cr & {T_8} = a + 7d = - 1\,......\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = - 3\,\& \,a = 20 \cr & {T_{14}} = a + 13d \cr & \,\,\,\,\,\,\,\,\,\, = 20 + 13\left( { - 3} \right) \cr & \,\,\,\,\,\,\,\,\,\, = - 19 \cr} $$

$$\eqalign{ & {T_2} = a + d = 8\,.......\,\left( 1 \right) \cr & {T_6} = a + 5d = 20\,......\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = 3\,\& \,a = 5 \cr & {T_{20}} = a + 19d \cr & \,\,\,\,\,\,\,\,\,\, = 5 + 19\left( 3 \right) \cr & \,\,\,\,\,\,\,\,\,\, = 62 \cr} $$

First term of AP = a = -20 and last term = l = 28 Number of terms = n = 17 $$\eqalign{ & {\text{Sum of AP}} = \frac{{\text{n}}}{2}\left( {{\text{a}} + {\text{l}}} \right) \cr & = \frac{{17}}{2}\left( { - 20 + 28} \right) \cr & = 17 \times 4 \cr & = 68 \cr} $$

Let the first term of an AP = a and the common difference = d 4 th term of AP = A 4 = a + 3d =11 ......(1) 7 th term = A 7 = a + 6d = -4 ......(2) Subtracting equation (1) from (2), we get : ⇒ 6d - 3d = -4 -11 ⇒ 3d = -15 ⇒ d = $$\frac{{ - 15}}{3}$$ = -5 Substituting it in equation (1) ⇒ a = 11 - 3(-5) = 11 + 15 = 26 ∴ 15 th term = A 15 = a + 14d = 26 + 14(-5) = 26 - 70 = -44

Let the first term of an AP = a and the common difference = d 3 rd term of AP = A 3 = a + 2d = -13 ...... (1) 8 th term = A 8 = a + 7d = 2 ...... (2) Subtracting equation (1) from (2), we get : ⇒ 7d - 2d = 2 - (-13) ⇒ 5d = 15 ⇒ d = $$\frac{{15}}{5}$$ = 3 Substituting it in equation (2) ⇒ a = 2 - 7(3) = 2 - 21 = -19 ∴ 14 th term = A 14 = a + 13d = -19 + 13(3) = -19 + 39 = 20

$$\eqalign{ & {T_3} = a + 2d = - 1.....\,\left( 1 \right) \cr & {T_8} = a + 7d = 19\,.....\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = 4\,\& \,a = - 9 \cr & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr} $$ $${S_{11}} = \frac{{11}}{2}$$ $$\left[ {2\left( { - 9} \right) + \left( {11 - 1} \right)\left( 4 \right)} \right]$$ $$\eqalign{ & {S_{11}} = \frac{{11}}{2}\left[ {\left( { - 18} \right) + \left( {40} \right)} \right] \cr & {S_{11}} = \frac{{11}}{2}\left[ {22} \right] \cr & {S_{11}} = 121 \cr} $$

$$\eqalign{ & {S_n} = \frac{n}{2}\left[ {a + l} \right] \cr & {S_{13}} = \frac{{13}}{2}\left[ { - 10 + 26} \right] \cr & {S_{13}} = \frac{{13}}{2}\left[ {16} \right] \cr & {S_{13}} = 104 \cr} $$

First term of AP = a = -19 and last term = l = 36 Number of terms = n = 12 $$\eqalign{ & {\text{Sum of AP}} = \frac{n}{2}\left( {a + l} \right) \cr & = \frac{{12}}{2}\left( { - 19 + 36} \right) \cr & = 17 \times 6 \cr & = 102 \cr} $$

$$\eqalign{ & {T_3} = a + 2d = - 9\,.....\,\left( 1 \right) \cr & {T_7} = a + 6d = 11\,.....\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = 5\,\& \,a = - 19 \cr & {T_{15}} = a + 14d \cr & \,\,\,\,\,\,\,\,\,\, = - 19 + 14\left( 5 \right) \cr & \,\,\,\,\,\,\,\,\,\, = 51 \cr} $$

$$\eqalign{ & {T_3} = a + 2d = - 13\,.....\,\left( 1 \right) \cr & {T_6} = a + 5d = - 4\,.....\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = 3\& a = - 19 \cr & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr} $$ $${S_{12}} = \frac{{12}}{2}$$ $$\left[ {2\left( { - 19} \right) + \left( {12 - 1} \right)\left( 3 \right)} \right]$$ $$\eqalign{ & {S_{12}} = \left( 6 \right)\left[ { - 38 + 33} \right] \cr & {S_{12}} = - 30 \cr} $$

$$\eqalign{ & {T_3} = a + 2d = 13\,.....\,\left( 1 \right) \cr & {T_5} = a + 4d = 21\,.....\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = 4\& a = 5 \cr & {T_{13}} = a + 12d \cr & \,\,\,\,\,\,\,\,\, = 5 + 12\left( 4 \right) \cr & \,\,\,\,\,\,\,\,\, = 5 + 48 \cr & \,\,\,\,\,\,\,\,\, = 53 \cr} $$

T 3 = a + 2d = 13 ...... (1) T 6 = a + 5d = -5 ...... (2) on solving (1) and (2) d = -6 & a = 25 T 11 = a + 10d = 25 + 10(-6) = -35

Let the first term of an AP = a and the common difference = d 3 th term of AP = A 3 = a + 2d = -8 ......(1) 9 th term = A 9 = a + 8d = 10 ...... (2) Subtracting equation (1) from (2), we get : ⇒ 8d - 2d = 10 - (-8) ⇒ 6d =18 ⇒ d = $$\frac{{18}}{6}$$ = 3 Substituting it in equation (2), ⇒ a = 10 - 8(3) = 10 - 24 = -14 ∴ 16 th term = A 16 = a + 15d = -14 + 15(3) = -14 + 45 = 31

Arithmetic mean of 7, 5, 13, x and 9 = 10 $$\eqalign{ & \Rightarrow \frac{{7 + 5 + 13 + x + 9}}{5} = 10 \cr & \Rightarrow 34 + x = 10 \times 5 \cr & \Rightarrow 34 + x = 50 \cr & \Rightarrow x = 50 - 34 \cr & \Rightarrow x = 16 \cr} $$

$$\eqalign{ & {S_n} = \frac{n}{2}\left( {a + l} \right) \cr & {S_9} = \frac{9}{2}\left( {7 + 55} \right) \cr & {S_9} = \frac{9}{2} \times 62 \cr & {S_9} = 9 \times 31 \cr & \,\,\,\,\,\,\,\,\, = 279 \cr} $$

Let the first term of the AP be a and the common difference = d 7 th term = A 7 = a + 6d 11 th term = A 11 = a + 10d According to question, ⇒ 7 × (a + 6d) = 11 × (a + 10d) ⇒ 7a + 42d = 11a + 110d ⇒ 11a - 7a = 42d - 110d ⇒ 4a = -68d ⇒ a = -17d ⇒ a + 17d = 0 = A 18

T 7 = a + 6d = -15 ....... (1) T 12 = a + 11d = 5 ...... (2) on solving (1) and (2) d = 4 & a = -39 T 16 = a + 15d = -39 + 15(4) = -39 + 60 = 21

a n = a + (n - 1) × d ⇒ a 25 = a + 24d and a 20 = a + 19d a 25 - a 20 = 45 ⇒ a + 24d - a - 19d = 45 ⇒ 5d = 45 ⇒ d = 9

T n = a + (n - 1) × d T 18 = a + 17d T 8 = a + 7d T 18 - T 8 = 17d - 7d = 10d

a n = a + (n - 1) × d where d = -3, Let a n = 0 ⇒ 0 = 24 + (n - 1) × -3 ⇒ 0 = 24 - 3n + 3 ⇒ 3n = 27 ⇒ n = 9 ⇒ 10 th term will be negative (-ve)

a n = a + (n - 1) × d where, d = x - 2 -x + 7 = 5, a = x - 7 ⇒ a 15 = (x - 7) + (15 - 1) × 5 ⇒ a 15 = (x - 7) + 14 × 5 ⇒ a 15 = x - 7 + 70 ⇒ a 15 = x + 63

$$\eqalign{ & {S_n} = \frac{1}{2}\left( {a + l} \right) \times n \cr & \Rightarrow 399 = \left( {1 + 20} \right) \times \frac{n}{2} \cr & \Rightarrow 399 \times 2 = 21 \times n \cr & \Rightarrow n = 399 \times \frac{2}{{21}} \cr & \Rightarrow n = 19 \times 2 \cr & \Rightarrow n = 38 \cr} $$

$${S_n} = \left[ {2a + \left( {n - 1} \right)d} \right] \times \frac{n}{2}$$ $$ \Rightarrow {S_5} = \left[ {2 \times 3 + \left( {5 - 1} \right)3} \right]$$ $$ \times \frac{5}{2}$$ $$\eqalign{ & \Rightarrow {S_5} = \left[ {6 + 12} \right] \times \frac{5}{2} \cr & \Rightarrow {S_5} = 18 \times \frac{5}{2} \cr & \Rightarrow {S_5} = 9 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 45 \cr} $$

$$\eqalign{ & {S_n} = \left[ {2a + \left( {n - 1} \right)d} \right] \times \frac{n}{2} \cr & {\text{Here,}} \cr & d = 1 \cr & a = 1\, \cr & {\text{and}}\,\,n - 1\,\,{\text{terms}} \cr} $$ $$ \Rightarrow {S_{n - 1}} = \left[ {2 + \left( {n - 1 - 1} \right)} \right]$$ $$ \times \frac{{\left( {n - 1} \right)}}{2}$$ $$ \Rightarrow {S_{n - 1}} = \left[ {2 + n - 2} \right]$$ $$ \times \frac{{\left( {n - 1} \right)}}{2}$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n\left( {n - 1} \right)}}{2}$$

a n = a + (n - 1) × d ⇒ 4 = a + (7 - 1) × -4 ⇒ 4 = a - 24 ⇒ a = 4 + 24 = 28

a n = a + (n - 1) × d where d = -4, Let a n = 0 ⇒ 0 = 92 + (n - 1) × -4 ⇒ 0 = 92 - 4n + 4 ⇒ 4n = 96 ⇒ n = 24

Let 1 st term = x = a + 1, 2 nd term = y = 2a + 1 and 3 rd term = z = 4a - 1 ⇒ y - x = z - y ⇒ 2y = x + z ⇒ 2(2a + 1) = a + 1 + 4a - 1 ⇒ 4a + 2 = 5a ⇒ a = 2