Progressions - Aptitude MCQ Questions and Solutions with Explanations

(2K - 1) - K = (2K + 1) - (2K - 1)
2K - 1 - K = 2
⇒ K = 3

First term of an A.P. (a) = 1
Last term (l) = 11
and sum of its terms = 36
Let n be the number of terms and d be the common difference, then
$$\eqalign{ & {a_n} = 1 = a + \left( {n - 1} \right)d = 11 \cr & \Rightarrow 1 + \left( {n - 1} \right)d = 11 \cr & \Rightarrow \left( {n - 1} \right)d = 11 - 1 \cr & \Rightarrow \left( {n - 1} \right)d = 10\,.....\,\left( 1 \right) \cr & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = 36 \cr} $$
$$ \Rightarrow \frac{n}{2}\left[ {2 \times 1 + 10} \right] = 36\,$$       $$\left[ {{\text{From}}\,\left( 1 \right)} \right]$$
$$\eqalign{ & \Rightarrow n\left( {2 + 10} \right) = 72 \cr & \Rightarrow 12n = 72 \cr & \Rightarrow n = \frac{{72}}{{12}} \cr & \Rightarrow n = 6 \cr} $$

S_n is the sum of n terms of an A.P.
a is its first term and d is common difference
$$\eqalign{ & d = {S_n} - k{S_{n - 1}} + {S_{n - 2}} \cr & \Rightarrow k{S_{n - 1}} = {S_n} + {S_{n - 2}} - d \cr & = \left( {{a_n} + {S_{n - 1}}} \right) + \left( {{S_{n - 1}} - {a_{n - 1}} - 1} \right) - d \cr} $$

\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l} {∵ S_n} = {S_{n - 1}} + {a_n}\,{\rm{and}}\\ {S_{n - 1}} = {a_{n - 1}} + {S_{n - 2}}\\ \Rightarrow {S_{n - 2}} = {S_{n - 1}} - {a_{n - 1}} \end{array} \right\}\]

$$\eqalign{ & = {a_n} + 2{S_{n - 1}} - {a_{n - 1}} - d \cr & = 2{S_{n - 1}} + {a_n} - {a_{n - 1}} - d \cr & = 2{S_{n - 1}} + d - d\,\,\left( {\because {a_n} - {a_{n - 1}} = d} \right) \cr & = 2{S_{n - 1}} \cr & \therefore k = 2 \cr} $$

S_n = n²p, S_m = m²p
∴ S_r = r²p and S_p = p²p = p³
Hence S_p = p³

The A.P. is 3, 7, 11, 15, ......
Where a = 3, d = 7 - 3 = 4 and sum S_n = 406
$$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & \Rightarrow 406 = \frac{n}{2}\left[ {2 \times 3 \times \left( {n - 1} \right) \times 4} \right] \cr & \Rightarrow 812 = n\left( {6 + 4n - 4} \right) \cr & \Rightarrow 812 = n\left( {4n + 2} \right) \cr & \Rightarrow 4{n^2} + 2n - 812 = 0 \cr & \Rightarrow 2{n^2} + n - 406 = 0 \cr & \Rightarrow 2{n^2} + 29n - 28n - 406 = 0 \cr & \Rightarrow n\left( {2n + 29} \right) - 14\left( {2n + 29} \right) = 0 \cr & \Rightarrow \left( {2n + 29} \right)\left( {n - 14} \right) = 0 \cr & \therefore n = 14\,{\text{or}}\,\frac{{ - 29}}{2} \cr & {\text{But}}\,n = \frac{{ - 29}}{2}\,{\text{is}}\,{\text{not}}\,{\text{possible}} \cr} $$

Sum of n terms = S_n
∴ a_n = S_n - S_{n - 1}
and a_{n - 1} = S_{n - 1} - S_{n - 2}
∴ Common difference (d) = a_n - a_{n - 1}
= (S_n - S_{n - 1}) - (S_{n - 1} - S_{n - 2})
S_n - S_{n - 1} - S_{n - 1} + S_{n - 2}
= S_n - 2S_{n - 1} + S_{n - 2}

In two A.P.’s common-difference is same
Let A and a are two A.P. ’s
First term of A is 8 and first term of a is 3
A₃₀ – a₃₀ = 8 + (30 – 1) d – 3 – (30 – 1) d
= 5 + 29d – 29d
= 5

$$\eqalign{ & {a_n} = 2n + 1 \cr & {a_1} = 2 \times 1 + \,1 = 2 + 1 = 3 \cr & {a_2} = 2 \times 2 + 1 = 4 + 1 = 5 \cr & \therefore d = {a_2} - {a_1} = 5 - 3 = 2 \cr & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = \frac{n}{2}\left[ {2 \times 3 + \left( {n - 1} \right) \times 2} \right] \cr & = \frac{n}{2}\left[ {6 + 2n - 2} \right] \cr & = \frac{n}{2}\left[ {2n + 4} \right] \cr & = n\left[ {n + 2} \right] \cr} $$

$$\eqalign{ & A.P.\,is\,\frac{1}{{2b}},\,\frac{{1 - 6b}}{{2b}},\,\frac{{1 - 12b}}{{2b}},.... \cr & \Rightarrow \frac{1}{{2b}},\,\frac{1}{{2b}} - \frac{{6b}}{{2b}},\,\frac{1}{{2b}} - \frac{{12b}}{{2b}},\,.... \cr & \Rightarrow \frac{1}{{2b}},\,\frac{1}{{2b}} - 3,\,\frac{1}{{2b}} - 6,\,.... \cr & \therefore d = \frac{1}{{2b}} - 3 - \frac{1}{{2b}} = - 3 \cr} $$

Sum of n terms of an A.P. = 3n² + n
and common difference (d) = 6
Let first term be a, then
$$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = 3{n^2} + n \cr & \Rightarrow \frac{n}{2}\left[ {2a + \left( {n - 1} \right)6} \right] = 3{n^2} + n \cr & \Rightarrow 2a + 6n - 6 = \left( {3{n^2} + n} \right) \times \frac{2}{n} \cr & \Rightarrow 2a + 6n - 6 = n\frac{{\left( {3n + 1} \right) \times 2}}{n} \cr & \Rightarrow 2a + 6n - 6 = \left( {3n + 1} \right)2 \cr & \Rightarrow 2a + 6n - 6 = 6n + 2 \cr & \Rightarrow 2a = 6n + 2 - 6n + 6 \cr & \Rightarrow 2a = 8 \cr & \therefore a = \frac{8}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 4 \cr} $$