Practice MCQ Questions and Answer on Progressions

Solution: $${S_n} = \left[ {2a + \left( {n - 1} \right)d} \right] \times \frac{n}{2}$$ $$ \Rightarrow {S_5} = \left[ {2 \times 3 + \left( {5 - 1} \right)3} \right]$$     $$ \times \frac{5}{2}$$ $$\eqalign{ & \Rightarrow {S_5} = \left[ {6 + 12} \right] \times \frac{5}{2} \cr & \Rightarrow {S_5} = 18 \times \frac{5}{2} \cr & \Rightarrow {S_5} = 9 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 45 \cr} $$

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

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

Solution: Answer & Solution Answer: Option E Solution: Common ratio, r = $$\frac{{20}}{5}$$ = 4 Last term or nth term of GP = arn - 1 20480 = 5 × (4n - 1) Or, 4n - 1 = $$\frac{{20480}}{5}$$  = 46 So, comparing the power, Thus, n - 1 = 6 Or, n = 7 Number of terms = 7

Solution: Let a1, d2 be the first terms of two ratios S and S' and d1, d2 be their common difference respectively Then, $$\eqalign{ & {S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]\,{\text{and}} \cr & S{'_n} = \frac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right] \cr & {\text{Now,}} \cr & \,\frac{{{S_n}}}{{S{'_n}}} = \frac{{\frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]}}{{\frac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right]}} \cr & = \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} \cr & {\text{But}}\,\frac{{{S_n}}}{{S{'_n}}} = \frac{{5n + 4}}{{9n + 6}} \cr & \therefore \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} = \frac{{5n + 4}}{{9n + 6}} \cr} $$ Now we have to find the ratios in 18th term Here n = 18 $$\eqalign{ & \therefore \frac{{2{a_1} + \left( {18 - 1} \right){d_1}}}{{2{a_2} + \left( {18 - 1} \right){d_2}}} = \frac{{5\left( {2n - 1} \right) + 4}}{{9\left( {2n - 1} \right) + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{5\left( {2 \times 18 - 1} \right) + 4}}{{9\left( {2 \times 18 - 1} \right) + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{5 \times 35 + 4}}{{9 \times 35 + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{175 + 4}}{{315 + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{179}}{{321}} \cr} $$

Solution: Sn is the sum of first n terms Last term nth term = Sn - Sn - 1

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

Solution: 1st Method: 8th term = a + 7d = 39 ........... (i) 12th term = a + 11d = 59 ........... (ii) (i) - (ii); Or, a + 7d - a - 11d = 39 - 59 Or, 4d = 20 Or, d = 5 Hence, a + 7 × 5 = 39 Thus, a = 39 - 35 = 4 2nd Method (Thought Process): 8th term = 39 And, 12th term = 59 Here, we see that 20 is added to 8th term 39 to get 12th term 59 i.e. 4 times the common difference is added to 39 So, CD = $$\frac{{20}}{4}$$  = 5 Hence, 7 times CD is added to 1st term to get 39. That means 4 is the 1st term of the AP

Solution: 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 = $$\frac{n}{2}$$ $$\eqalign{ & {S_2} = \frac{n}{{2 \times 2}}\left[ {2 \times 3 + \left( {\frac{n}{2} - 1} \right) \times 4} \right] \cr & \,\,\,\,\,\,\,\,\, = \frac{n}{4}\left[ {6 + 2n - 4} \right] \cr & \,\,\,\,\,\,\,\,\, = \frac{n}{4}\left[ {2n + 2} \right] \cr & \,\,\,\,\,\,\,\,\, = \frac{{n\left( {n + 1} \right)}}{2} \cr & \therefore \frac{{{s_1}}}{{{s_2}}} = \frac{{{n^2} \times 2}}{{n\left( {n + 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2n}}{{n + 1}} \cr} $$

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