Progressions - Aptitude MCQ Questions and Solutions with Explanations

Let the cost of an equipment is Rs. 100
Now the percentages of depreciation at the end of 1st, 2^nd, 3^rd years are 15, 13.5, 12, which are in A.P., with a = 15 and d = - 1.5
Hence, percentage of depreciation in the tenth year = a + (10 - 1) d = 15 + 9 (-1.5) = 1.5
Also total value depreciated in 10 years = 15 + 13.5 + 12 + ..... + 1.5 = 82.5
Hence, the value of equipment at the end of 10 years = 100 - 82.5 = 17.5
The total cost being
$$\eqalign{ & = {\text{Rs}}{\text{.}}\,\frac{{6,00,000}}{{100}} \times 17.5 \cr & = {\text{Rs}}{\text{.}}\,1,05,000 \cr} $$

First term is -64. The common difference is -2. The last term is -100.
Sum of the first n terms of an AP =
$$\frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]$$
To compute the sum, we know the first term a1 = -64 and the common difference d = -2.
We do not know the number of terms n. Let us first compute the number of terms and then find the sum of the terms.
a_n = a₁ + (n - 1)d
-100 = -64 + (n - 1)(-2)
Therefore, n = 19
Sum =
$${S_n} = \frac{{19}}{2}$$ $$\left[ {2\left( { - 64} \right) + \left( {19 - 1} \right)\left( { - 2} \right)} \right]$$
$$\eqalign{ & {S_n} = \frac{{19}}{2}\left[ { - 128 - 36} \right] \cr & {S_n} = 19 \times \left( { - 82} \right) \cr & {S_n} = - 1558 \cr} $$

The positive integers, which are divisible by 5 are 5, 10, 15, ....., 1000
Out of these 10, 20, 30, ......, 1000 are divisible by 2
Thus, we have to find the sum of the positive integers 5, 15, 25, ......, 995
If n is the number of terms in it the sequence then
995 = 5 + 10(n - 1)
⇒ 1000 = 10n
∴ n = 100
Thus the sum of the series
$$\eqalign{ & = \left( {\frac{n}{2}} \right)\left( {a + l} \right) \cr & = \left( {\frac{{100}}{2}} \right)\left( {5 + 995} \right) \cr & = \frac{{100 \times 1000}}{2} \cr & = 50000 \cr} $$

Sum of n terms of an A.P. = 3n² + 5n
Let a be the first term and d be the common difference
S_n = 3n² + 5n
S₁ = 3(1)² + 5 × 1 = 3 + 5 = 8
S₂ = 3(2)² + 5 × 2 = 12 + 10 = 22
∴ First term (a) = 8
a₂ = S₂ - S₁ = 22 - 8 = 14
d = a₂ - a₁ = 14 - 8 = 6
Now a_n = a + (n - 1)d
⇒ 164 = 8 + (n - 1) × 6
⇒ 6n - 6 = 164 - 8
⇒ 6n = 156 + 6
⇒ 6n = 162
⇒ n = $$\frac{{162}}{6}$$
⇒ n = 27
∴ 168 is 27^th term

$$\eqalign{ & S = \frac{n}{2}\left( {l + a} \right) \cr & l = a + \left( {n - 1} \right)d \cr & d = \frac{{{l^2} - {a^2}}}{{k - \left( {l + a} \right)}}\,{\text{and}}\,{\text{also}}\,d = \frac{{l - a}}{{n - 1}} \cr & \therefore \frac{{l - a}}{{n - 1}} = \frac{{\left( {l + a} \right)\left( {l - a} \right)}}{{k - \left( {l + a} \right)}} \cr & \Rightarrow \frac{1}{{n - 1}} = \frac{{l + a}}{{k - \left( {l + a} \right)}} \cr & \Rightarrow k - \left( {l + a} \right) = \left( {n - 1} \right)\left( {l + a} \right) \cr & \Rightarrow k = \left( {n - 1} \right)\left( {l + a} \right) + \left( {l + a} \right) \cr & \Rightarrow k = \left( {l + a} \right)\left( {n - 1 + 1} \right) \cr & \Rightarrow k = n\left( {l + a} \right) \cr & \, \Rightarrow k = 2 \times \frac{n}{2}\left( {l + a} \right)\left\{ {\therefore \frac{n}{2}\left( {l + a} \right) = S} \right\} \cr & \Rightarrow k = 2 \times S \cr & \Rightarrow k = 2S \cr} $$

$${S_n}$$ = Sum of $$n$$ terms of A.P. and $${S_{2n}}$$ = $$3{S_n}$$
$${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right],$$     $${S_{2n}} = \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right]$$     and $${S_{3n}} = \frac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]$$
We know that $${S_{3n}} = 3\left( {{S_{2n}} - {S_n}} \right)$$    and $${S_{2n}} = 3{S_n}$$
$$\eqalign{ & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{{3\left( {{S_{2n}} - {S_n}} \right)}}{{{S_n}}} \cr & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{{3\left( {3{S_n} - {S_n}} \right)}}{{{S_n}}} \cr & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{{3 \times 2{S_n}}}{{{S_n}}} \cr & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{6}{1} \cr & \therefore {S_{3n}}:{S_n} = 6 \cr} $$

The series is given
$$\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} $$     $$ + $$ ......
$$ \Rightarrow \sqrt 2 + 2\sqrt 2 + 3\sqrt 2 $$     $$ + $$ $$4\sqrt 2 $$  $$ + $$ ......
Here a = $$\sqrt 2 $$   and d = $$2\sqrt 2 $$  $$ - $$ $$\sqrt 2 $$  = $$\sqrt 2 $$
$$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = \frac{n}{2}\left[ {2\sqrt 2 + \left( {n - 1} \right)\sqrt 2 } \right] \cr & = \frac{n}{2}\left[ {2\sqrt 2 + \sqrt 2 \,n - \sqrt 2 } \right] \cr & = \frac{n}{2}\left( {\sqrt 2 \,n + \sqrt 2 } \right) \cr & = \frac{{n\sqrt 2 }}{2}\left( {n + 1} \right) \cr & = \frac{{n\left( {n + 1} \right)}}{{\sqrt 2 }} \cr} $$

In first A.P. let its first term be a₁ and common difference d₁
and in second A.P., first term be a₂ and common difference d₂, then
$$\eqalign{ & \frac{{{S_n}}}{{{S_n}}} = \frac{{\frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]}}{{\frac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right]}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} \cr} $$
$$\therefore \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} = \frac{{3n + 5}}{{5n + 7}}$$
$${\text{Substituting n = 2n - 1, then}}$$
$$\frac{{2{a_1} + \left( {2n - 2} \right){d_1}}}{{2{a_2} + \left( {2n - 2} \right){d_2}}} = $$     $$\frac{{3\left( {2n - 1} \right) + 5}}{{5\left( {2n - 1} \right) + 7}}$$
$$ \Rightarrow \frac{{{a_1} + \left( {n - 1} \right){d_1}}}{{{a_2} + \left( {n - 1} \right){d_2}}} = $$     $$\frac{{6n - 3 + 5}}{{10n - 5 + 7}}$$    (Dividing by 2)
$$\eqalign{ & \Rightarrow \frac{{{a_{1n}}}}{{{a_{2n}}}} = \frac{{6n + 2}}{{10n + 2}} \cr & \Rightarrow \frac{{{a_{1n}}}}{{{a_{2n}}}} = \frac{{3n + 1}}{{5n + 1}} \cr} $$

18, a, b - 3 are in A.P.,
then a - 18 = -3 - b
⇒ a + b = -3 + 18 = 15

$$\eqalign{ & {18^{{\text{th}}}}\,{\text{term}}:{11^{{\text{th}}}}\,{\text{term}} = 3:2 \cr & \Rightarrow \frac{{{a_{18}}}}{{{a_{11}}}} = \frac{3}{2} \cr & \Rightarrow \frac{{a + 17d}}{{a + 10d}} = \frac{3}{2} \cr & \Rightarrow 2a + 34d = 3a + 30d \cr & \Rightarrow 34d - 30d = 3a - 2a \cr & \Rightarrow a = 4d \cr & {\text{Now,}} \cr & \frac{{{a_{21}}}}{{{a_5}}} = \frac{{a + 20d}}{{a + 4d}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{4d + 20d}}{{4d + 4d}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{24d}}{{8d}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{3}{1} \cr & \therefore {a_{21}}:{a_5} = 3:1 \cr} $$