Practice MCQ Questions and Answer on Progressions

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: an = a + (n - 1) × d where d = -3, Let an = 0 ⇒ 0 = 24 + (n - 1) × -3 ⇒ 0 = 24 - 3n + 3 ⇒ 3n = 27 ⇒ n = 9 ⇒ 10th term will be negative (-ve)

Solution: $$\eqalign{ & {T_3} = a + 2d = 13\,.....\,\left( 1 \right) \cr & {T_5} = a + 4d = 21\,.....\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = 4\& a = 5 \cr & {T_{13}} = a + 12d \cr & \,\,\,\,\,\,\,\,\, = 5 + 12\left( 4 \right) \cr & \,\,\,\,\,\,\,\,\, = 5 + 48 \cr & \,\,\,\,\,\,\,\,\, = 53 \cr} $$

Solution: In an A.P. Sp = q, Sq = p Sp+q = Sum of (p + q) terms = Sum of p term + Sum of q terms = p + q

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: We will it through option checking method: $$\eqalign{ & {\frac{5}{9}} \times \left( {{{10}^n} - 1} \right) \cr & {\text{We}}{\kern 1pt} {\kern 1pt} {\text{put}}{\kern 1pt} {\kern 1pt} n = 1, \cr & {\frac{5}{9}} \times \left( {{{10}^1} - 1} \right) = 5 \cr & n = 2\left( {\frac{5}{9}} \right) \times \left( {{{10}^2} - 1} \right) = 55 \cr & n = 3\left( {\frac{5}{9}} \right) \times \left( {{{10}^3} - 1} \right) = 555 \cr} $$ It means Option C is satisfying the sequence so the nth term would be $${\kern 1pt} {\frac{5}{9}} \times \left( {{{10}^n} - 1} \right)$$

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

Solution: T3 = a + 2d = 13 ...... (1) T6 = a + 5d = -5 ...... (2) on solving (1) and (2) d = -6 & a = 25 T11 = a + 10d       = 25 + 10(-6)       = -35

Solution: 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 $$\eqalign{ & = \left( {\frac{n}{2}} \right)\left( {a + l} \right) \cr & = \left( {\frac{{100}}{2}} \right)\left( {5 + 995} \right) \cr & = \frac{{100 \times 1000}}{2} \cr & = 50000 \cr} $$

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