Progressions - Aptitude MCQ Questions and Solutions with Explanations

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 = A₄ = a + 3d =11 ......(1)
7^th term = A₇ = 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 = A₁₅ = 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 = A₃ = a + 2d = -13 ...... (1)
8^th term = A₈ = 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 = A₁₄ = a + 13d
= -19 + 13(3)
= -19 + 39
= 20

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

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