Progressions — Part 2 MCQs 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 = a 1 + (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. = 3n 2 + 5n Let a be the first term and d be the common difference S n = 3n 2 + 5n S 1 = 3(1) 2 + 5 × 1 = 3 + 5 = 8 S 2 = 3(2) 2 + 5 × 2 = 12 + 10 = 22 ∴ First term (a) = 8 a 2 = S 2 - S 1 = 22 - 8 = 14 d = a 2 - a 1 = 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 a 1 and common difference d 1 and in second A.P., first term be a 2 and common difference d 2 , 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

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

(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 = n 2 p, S m = m 2 p ∴ S r = r 2 p and S p = p 2 p = p 3 Hence S p = p 3

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 A 30 – a 30 = 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. = 3n 2 + n and common difference (d) = 6 Let first term be a, then $$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = 3{n^2} + n \cr & \Rightarrow \frac{n}{2}\left[ {2a + \left( {n - 1} \right)6} \right] = 3{n^2} + n \cr & \Rightarrow 2a + 6n - 6 = \left( {3{n^2} + n} \right) \times \frac{2}{n} \cr & \Rightarrow 2a + 6n - 6 = n\frac{{\left( {3n + 1} \right) \times 2}}{n} \cr & \Rightarrow 2a + 6n - 6 = \left( {3n + 1} \right)2 \cr & \Rightarrow 2a + 6n - 6 = 6n + 2 \cr & \Rightarrow 2a = 6n + 2 - 6n + 6 \cr & \Rightarrow 2a = 8 \cr & \therefore a = \frac{8}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 4 \cr} $$

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

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

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

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 (a 7 ) = a + 6d = 34 13 th term (a 13 ) = a + 12d = 64 Subtracting, 6d = 30 ⇒ d = 5 and a + 12 x 5 = 64 ⇒ a + 60 = 64 ⇒ a = 64 – 60 = 4 18 th term (a 18 ) = 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 = n 2 $$\eqalign{ & \therefore n\left( {n + 1} \right) = k{n^2} \cr & \Rightarrow k = \frac{{n(n + 1)}}{{{n^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n + 1}}{n} \cr} $$

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