Progressions - Aptitude MCQ Questions and Solutions with Explanations

4 numbers are in A.P.
Let the numbers be
a – 3d, a – d, a + d, a + 3d
Where a is the first term and 2d is the common difference
Now their sum = 50
a – 3d + a – d + a + d + a + 3d = 50
and greatest number is 4 times the least number
a + 3d = 4 (a – 3d)
a + 3d = 4a – 12d
4a – a = 3d + 12d
⇒ 3a = 15d
$$\eqalign{ & \Rightarrow a = \frac{{15d}}{3} = 5d \cr & \Rightarrow \frac{{25}}{2} = 5d \cr & \Rightarrow d = \frac{{25}}{{2 \times 5}} \cr & \Rightarrow d = \frac{5}{2} \cr & \therefore {\text{Numbers}}\,{\text{are}} \cr} $$
$$\frac{{25}}{2} - 3 \times \frac{5}{2},$$   $$\,\frac{{25}}{2} - \frac{5}{2},$$   $$\frac{{25}}{2} + \frac{5}{2},$$   $$\frac{{25}}{2} + 3 \times \frac{5}{2}$$
$$\eqalign{ & \Rightarrow \frac{{10}}{2},\,\frac{{20}}{2},\,\frac{{30}}{2},\,\frac{{40}}{2} \cr & \Rightarrow 5,\,10,\,15,\,20 \cr} $$

Odd numbers are 1, 3, 5, 7, 9, 11, 13, ...... n
∴ S₁ = Sum of odd numbers = n²
S₂ = Sum of number at odd places
3, 7, 11, 15, ......
a = 3, d = 7 - 3 = 4 and number of term = $$\frac{n}{2}$$
$$\eqalign{ & {S_2} = \frac{n}{{2 \times 2}}\left[ {2 \times 3 + \left( {\frac{n}{2} - 1} \right) \times 4} \right] \cr & \,\,\,\,\,\,\,\,\, = \frac{n}{4}\left[ {6 + 2n - 4} \right] \cr & \,\,\,\,\,\,\,\,\, = \frac{n}{4}\left[ {2n + 2} \right] \cr & \,\,\,\,\,\,\,\,\, = \frac{{n\left( {n + 1} \right)}}{2} \cr & \therefore \frac{{{s_1}}}{{{s_2}}} = \frac{{{n^2} \times 2}}{{n\left( {n + 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2n}}{{n + 1}} \cr} $$

$$\eqalign{ & {\text{In}}\,{\text{an}}\,{\text{A}}{\text{.P}}{\text{.}} \cr & a = 2\,{\text{and}}\,d = 4,\,n = 40 \cr & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = \frac{{40}}{2}\left[ {2 \times 2 + \left( {40 - 1} \right) \times 4} \right] \cr & = 20\left[ {4 + 39 \times 4} \right] \cr & = 20 \times \left( {4 + 156} \right) \cr & = 20 \times 160 \cr & = 3200 \cr} $$

S_n is the sum of first n terms
Last term n^th term = S_n - S_{n - 1}

1, 3, 5, 7, ........ are n odd numbers
Where a = 1, and d = 2
$$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right) \times 2} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{n}{2}\left[ {2 + 2n - 2} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{n}{2} \times 2n \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {n^2} \cr} $$

$$\eqalign{ & {\text{Sum}}\,{\text{of}}\,n\,{\text{terms}}\left( {{S_n}} \right) = 3{n^2} + 5n \cr & \therefore {\text{Sum}}\,{\text{of}}\,\left( {n - 1} \right)\,{\text{terms}}\,\left( {{S_{n - 1}}} \right) \cr & = 3{\left( {n - 1} \right)^2} + 5\left( {n - 1} \right) \cr & = 3\left( {{n^2} - 2n + 1} \right) + 5n - 5 \cr & = 3{n^2} - 6n + 3 + 5n - 5 \cr & = 3{n^2} - n - 2 \cr & \therefore {n^{th}}\,{\text{term}} = {S_n} - {S_{n - 1}} \cr & \Rightarrow {a_n} = 3{n^2} + 5n - 3{n^2} + n + 2 \cr & \,\,\,\,\,\,\,\,\,\,{a_n} = 6n + 2,\,{\text{But}}\,{a_n} = 164 \cr & \Rightarrow 6n + 2 = 164 \cr & \Rightarrow 6n = 164 - 2 \cr & \Rightarrow 6n = 162 \cr & \therefore n = \frac{{162}}{6} = {27^{th}}\,{\text{term}} \cr} $$

$$\eqalign{ & {\text{A}}{\text{.P}}{\text{.is}}\,\,\frac{1}{3},\,\frac{{1 - 3b}}{3},\,\frac{{1 - 6b}}{3},\,...... \cr & \Rightarrow \frac{1}{3},\,\frac{1}{3} - \frac{{3b}}{3},\,\frac{1}{3} - \frac{{6b}}{3},\,...... \cr & \Rightarrow \frac{1}{3},\,\frac{1}{3} - b,\,\frac{1}{3} - 2b,\,....... \cr & \therefore d = \left( {\frac{1}{3} - b} \right) - \frac{1}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{3} - b - \frac{1}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = - b \cr} $$

$$\eqalign{ & {\text{Sum}}\,{\text{of}}\,p\,{\text{terms}} = q \cr & i.e., \cr & {S_p} = \frac{p}{2}\left[ {2a + \left( {p - 1} \right)d} \right] = q \cr & \Rightarrow p\left[ {2a + \left( {p - 1} \right)d} \right] = 2q \cr & \Rightarrow 2ap + p\left( {p - 1} \right)d = 2q\,.....\left( 1 \right) \cr & {\text{and}}\,{\text{sum}}\,{\text{of}}\,q\,{\text{terms}} = p \cr & i.e., \cr & {S_q} = \frac{q}{2}\left[ {2a + \left( {q - 1} \right)d} \right] = p \cr & \Rightarrow q\left[ {2a + \left( {q - 1} \right)d} \right] = 2p \cr & \Rightarrow 2aq + q\left( {q - 1} \right)d = 2p\,.....\left( 2 \right) \cr & {\text{Subtracting}}\,\left( 2 \right)\,{\text{from}}\,\left( 1 \right) \cr} $$
$$ \Rightarrow 2a\left( {p - q} \right) + \left\{ {{p^2} - p - {q^2} + q} \right\}d$$       $$ = 2q - 2p$$
$$ \Rightarrow 2a\left( {p - q} \right) + \left\{ {{p^2} - {q^2} - \left( {p - q} \right)} \right\}d$$       $$ = - 2\left( {p - q} \right)$$
$$ \Rightarrow 2a\left( {p - q} \right) + \left\{ {\left( {p + q} \right)\left( {p - q} \right) - \left( {p - q} \right)} \right\}d$$         $$ = - 2\left( {p - q} \right)$$
$$ \Rightarrow 2a\left( {p - q} \right) + \left( {p - q} \right)\left[ {p + q - 1} \right]d$$       $$ = - 2\left( {p - q} \right)$$   $${\text{Dividing}}\,{\text{by}}$$   $$\,\left( {p - q} \right)$$
$$\eqalign{ & \Rightarrow 2a + \left( {p + q - 1} \right)d = - 2\,.....\left( 3 \right) \cr & {S_{p + q}} = \frac{{p + q}}{2}\left[ {2a + \left( {p + q - 1} \right)d} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{p + q}}{2}\left[ { - 2} \right]\,\,\,\left[ {{\text{From}}\,\left( 3 \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = - \left( {p + q} \right) \cr} $$

Let three consecutive terms of an increasing A.P. be a - d, a + d where a is the first term and d be the common difference
$$\eqalign{ & \therefore a - d + a + a + d = 51 \cr & \Rightarrow 3a + 51 \cr & \therefore a = \frac{{51}}{3} = 17 \cr} $$
and product of the first and third terms
$$\eqalign{ & = \left( {a - d} \right)\left( {a + d} \right) = 273 \cr & \Rightarrow {a^2} - {d^2} = 273 \cr & \Rightarrow {\left( {17} \right)^2} - {d^2} = 273 \cr & \Rightarrow 289 - {d^2} = 273 \cr & \Rightarrow {d^2} = 289 - 273 \cr & \Rightarrow {d^2} = 16 \cr & \Rightarrow {d^2} = {\left( { \pm 4} \right)^2} \cr & \therefore d = \pm 4 \cr & \because {\text{The A}}{\text{.P}}{\text{.}}\,{\text{is}}\,{\text{increasing}} \cr & \therefore d = 4 \cr & {\text{Now}}\,{\text{third}}\,{\text{term}} = a + d \cr & = 17 + 4 = 21 \cr} $$

First term (a₁) = a
Second term (a₂) = b
and last term (l) = 2a
∴ d = Second term - First term = b - a
∴ l = a_n = a + (n - 1)d
$$\eqalign{ & \Rightarrow 2a = a + \left( {n - 1} \right)\left( {b - a} \right) \cr & \Rightarrow \left( {n - 1} \right)\left( {b - a} \right) = a \cr & \Rightarrow n - 1 = \frac{a}{{b - a}} \cr & \Rightarrow n = \frac{a}{{b - a}} + 1 \cr & \Rightarrow n = \frac{{a + b - a}}{{b - a}} \cr & \Rightarrow n = \frac{b}{{b - a}} \cr & \therefore {S_n} = \frac{n}{2}\left[ {a + l} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{b}{{2\left( {b - a} \right)}}\left[ {a + 2a} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3ab}}{{2\left( {b - a} \right)}} \cr} $$