Progressions - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right], \cr & {S_{2n}} = \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right]\,{\text{and}} \cr & {S_{3n}} = \frac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right] \cr & {\text{Now}}\,{S_{2n}} - {S_n} \cr} $$
$$ = \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right] - $$     $$\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$
$$ = \frac{n}{2}\left[ {4a + \left( {4n - 2} \right)d} \right] - $$     $$\left[ {2a + \left( {n - 1} \right)d} \right]$$
$$\eqalign{ & = \frac{n}{2}\left[ {4a - 2a + \left( {4n - 2 - n + 1} \right)d} \right] \cr & = \frac{n}{2}\left[ {2a + \left( {3n - 1} \right)d} \right] \cr & = \frac{1}{3}\left( {{S_{3n}}} \right) \cr & \therefore {S_{3n}}:\left( {{S_{2n}} - {S_n}} \right) \cr & = 3:1\,{\text{or}}\,\frac{3}{1} = 3 \cr} $$

$$\eqalign{ & \frac{1}{{x + 2}},\,\frac{1}{{x + 3}},\,\frac{1}{{x + 5}}\,{\text{are}}\,{\text{in}}\,{\text{A}}{\text{.P}}{\text{.}} \cr & \therefore \frac{1}{{x + 3}} - \frac{1}{{x + 2}} = \,\frac{1}{{x + 5}} - \frac{1}{{x + 3}} \cr & \Rightarrow \frac{{x + 2 - x - 3}}{{\left( {x + 3} \right)\left( {x + 2} \right)}} = \frac{{x + 3 - x - 5}}{{\left( {x + 5} \right)\left( {x + 3} \right)}} \cr & \Rightarrow \frac{{ - 1}}{{\left( {x + 3} \right)\left( {x + 2} \right)}} = \frac{{ - 2}}{{\left( {x + 5} \right)\left( {x + 3} \right)}} \cr & \Rightarrow \frac{{ - 1}}{{x + 2}} = \frac{{ - 2}}{{x + 5}} \cr & \Rightarrow - 2x - 4 = - x - 5 \cr & \Rightarrow - 2x + x = - 5 + 4 \cr & \Rightarrow - x = - 1 \cr & \therefore x = 1 \cr} $$

In the given A.P.
First term = a and n^th term = b
$$\eqalign{ & \therefore a + \left( {n - 1} \right)d = b \cr & \Rightarrow \left( {n - 1} \right)d = b - a \cr & \Rightarrow d = \frac{{b - a}}{{n - 1}} \cr} $$

Sum of 5 + 9 + 13 + . . . . to n terms
$$\eqalign{ & = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & {\text{Here}}\,a = 5,\,d = 9 - 5 = 4 \cr & \therefore {\text{Sum}} = \frac{n}{2}\left[ {2 \times 5 + \left( {n - 1} \right) \times 4} \right] \cr & = \frac{n}{2}\left[ {10 + 4n - 4} \right] \cr & = \frac{n}{2}\left[ {6 + 4n} \right] \cr & = n\left( {3 + 2n} \right) \cr} $$
and sum of 7 + 9 + 11 + . . . . to (n + 1) terms
$$\eqalign{ & = \frac{{n + 1}}{2}\left[ {2 \times 7 + \left( {n + 1 - 1} \right)2} \right] \cr & = \frac{{n + 1}}{2}\left[ {14 + 2n} \right] \cr & = \left( {n + 1} \right)\left( {7 + n} \right) \cr & \therefore \frac{{5 + 9 + 13 + ...\,{\text{to}}\,n\,{\text{terms}}}}{{7 + 9 + 11 + ...\,{\text{to}}\,\left( {n + 1} \right)\,{\text{terms}}}} = \frac{{17}}{{16}} \cr & \Rightarrow \frac{{n\left( {3 + 2n} \right)}}{{\left( {n + 1} \right)\left( {7 + n} \right)}} = \frac{{17}}{{16}} \cr & \Rightarrow 16n\left( {3 + 2n} \right) = 17\left( {n + 1} \right)\left( {7 + n} \right) \cr & \Rightarrow 48n + 32{n^2} = 17\left( {{n^2} + 8n + 7} \right) \cr & \Rightarrow 48n + 32{n^2} = 17{n^2} + 136n + 119 \cr & \Rightarrow 48n + 32{n^2} - 17{n^2} - 136n - 119 = 0 \cr & \Rightarrow 15{n^2} - 88n - 119 = 0 \cr & \Rightarrow 15{n^2} - 105n + 17n - 119 = 0 \cr} $$

\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l} ∵ 15 \times \left( { - 119} \right) = 1785\\ - 1785 = 17 \times \left( {105} \right)\\ - 88 = 17 - 105 \end{array} \right\}\]

$$\eqalign{ & \Rightarrow 15n\left( {n - 7} \right) + 17\left( {n - 7} \right) = 0 \cr & \Rightarrow \left( {n - 7} \right)\left( {15n + 17} \right) = 0 \cr} $$
Either $$n - 7 = 0,$$   then $$n = 7$$
or $$15n + 13 = 0,$$    then $$n = \frac{{ - 13}}{{15}}$$   which is not possible being fraction
∴ n = 7

$$\eqalign{ & {\text{A}}{\text{.P}}{\text{.}}\,{\text{is}}\,\frac{1}{{2q}},\,\frac{{1 - 2q}}{{2q}},\,\frac{{1 - 4q}}{{2q}},.... \cr & \Rightarrow \frac{1}{{2q}},\,\left( {\frac{1}{{2q}} - 1} \right),\,\left( {\frac{1}{{2q}} - 2} \right),\,.... \cr & {\text{Clearly}}\,d = \left( {\frac{1}{{2q}} - 1} \right) - \frac{1}{{2q}} \cr & = \frac{1}{{2q}} - 1 - \frac{1}{{2q}} \cr & = - 1 \cr} $$

2 (3y + 5) = 3y – 1 + 5y + 1
(If a, b, c are in A.P., b – a = c – b ⇒ 2b = a + c)
⇒ 6y + 10 = 8y
⇒ 10 = 2y
⇒ y = 5

7^th term (a₇) = a + 6d = 34
13^th term (a₁₃) = a + 12d = 64
Subtracting, 6d = 30 ⇒ d = 5
and a + 12 x 5 = 64
⇒ a + 60 = 64
⇒ a = 64 – 60 = 4
18^th term (a₁₈)
= a + 17d
= 4 + 17 x 5
= 4 + 85
= 89

Let a be the first term and d be the common difference of an A.P. and
$$\eqalign{ & {S_n} = 2{n^2} + 5n \cr & \therefore {S_1} = 2{\left( 1 \right)^2} + 5 \times 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 2 + 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 7 \cr & \therefore {S_2} = 2{\left( 2 \right)^2} + 5 \times 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8 + 10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 18 \cr & \therefore {\text{First}}\,{\text{term}}\,\left( {{a_1}} \right) = 7\,{\text{and}} \cr & {\text{Second}}\,{\text{term}}\,{a_2} = {S_2} - {S_1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 18 - 7 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 11 \cr & \therefore d = {a_2} - {d_1} \cr & \,\,\,\,\,\,\,\,\,\,\, = 11 - 7 \cr & \,\,\,\,\,\,\,\,\,\,\, = 4 \cr & Now\,{a_n} = a + \left( {n - 1} \right)d \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7 + \left( {n - 1} \right)4 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7 + 4n - 4 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4n + 3 \cr} $$

Sum of n even natural number = n(n+1)
and sum of n odd natural numbers = n²
$$\eqalign{ & \therefore n\left( {n + 1} \right) = k{n^2} \cr & \Rightarrow k = \frac{{n(n + 1)}}{{{n^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n + 1}}{n} \cr} $$

In an A.P. S_p = q, S_q = p
S_p+q = Sum of (p + q) terms
= Sum of p term + Sum of q terms
= p + q