Sequences and Series MCQ Questions & Answers in Algebra | Maths

Given series $$63 + 65 + 67 + 69 \,. . . .\,\,\,\, …\left( {\text{i}} \right)$$
and $$3 + 10 + 17 + 24 \,. . . . \,\,\,\, …\left( {\text{ii}} \right)$$
Now from (i), $$m^{th}$$ term $$ = \left( {2m + 61} \right)$$
and $$m^{th}$$ term of (ii) series $$ = \left( {7m - 4} \right)$$
Under condition
⇒ $$7m - 4 = 2m + 61$$
⇒ $$5m = 65$$
⇒ $$m = 13.$$

$$\eqalign{ & {\text{Here }}\frac{1}{{{1^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + .....\,{\text{to }}\infty = \frac{{{\pi ^2}}}{8}. \cr & {\text{Let}}\,\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + .....\,{\text{to }}\infty = x. \cr & {\text{Then }}\,x = \frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + .....\,{\text{to }}\infty \cr & = \left( {\frac{1}{{{1^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + .....\,{\text{to }}\infty } \right) + \left( {\frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{6^2}}} + .....\,{\text{to }}\infty } \right) \cr & = \frac{{{\pi ^2}}}{8} + \frac{1}{4}\left( {\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + .....\,{\text{to }}\infty } \right) \cr & = \frac{{{\pi ^2}}}{8} + \frac{1}{4}x. \cr} $$

$${{n^{th}}}$$ term of the given series
$$\eqalign{ & = \,\,{T_n} = {\left( {n - 1} \right)^2} + \left( {n - 1} \right)n + {n^2} \cr & = \,\,\frac{{\left( {{{\left( {n - 1} \right)}^3} - {n^3}} \right)}}{{\left( {n - 1} \right) - n}} \cr & = \,\,{n^3} - {\left( {n - 1} \right)^3} \cr & \Rightarrow \,\,{S_n} = \sum\limits_{k = 1}^n {\left[ {{k^3} - {{\left( {k - 1} \right)}^3}} \right]} \cr & \Rightarrow \,\,8000 = {n^3} \cr & \Rightarrow \,\,n = 20{\text{ which is a natural number}}{\text{.}} \cr & {\text{Now, put }}n = 1,2,3,.....,20 \cr & {T_1} = {1^3} - {0^3} \cr & {T_2} = {2^3} - {1^3} \cr & \cdot \cr & \cdot \cr & \cdot \cr & {T_{20}} = {20^3} - {19^3} \cr & {\text{Now, }}{T_1} + {T_2} + - - - + {T_{20}} = {S_{20}} \cr & \Rightarrow \,\,{S_{20}} = {20^3} - {0^3} = 8000 \cr & {\text{Hence, both the given statements are true and}} \cr & {\text{statement 2 supports statement 1}}{\text{.}} \cr} $$

The series is
$$\eqalign{ & \left( {{x^2} + {x^4} + {x^6} + \,{\text{upto }}n{\text{ terms}}} \right) + \left( {\frac{1}{{{x^2}}} + \frac{1}{{{x^4}}} + \frac{1}{{{x^6}}} + {\text{upto }}n{\text{ terms}}} \right) + 2 + \left. {{\text{upto }}n{\text{ terms}}} \right) \cr & = \frac{{{x^2}\left( {{x^{2n}} - 1} \right)}}{{{x^2} - 1}} + \frac{{\frac{1}{{{x^2}}}\left( {1 - \frac{1}{{{x^{2n}}}}} \right)}}{{1 - \frac{1}{{{x^2}}}}} + 2n \cr & = \frac{{{x^2}\left( {{x^{2n}} - 1} \right)}}{{{x^2} - 1}} + \frac{{{x^{2n}} - 1}}{{\left( {{x^2} - 1} \right){x^{2n}}}} + 2n \cr & = \frac{{{x^{2n}} - 1}}{{{x^2} - 1}} \times \frac{{{x^{2n + 2}} + 1}}{{{x^{2n}}}} + 2n \cr} $$

Let $$a, b$$  and $$c$$ are in H.P.
$$\eqalign{ & \therefore \frac{1}{a},\frac{1}{b},\frac{1}{c}\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & \therefore \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \cr & \Rightarrow \frac{1}{c} = \frac{2}{b} - \frac{1}{a} \cr & {\text{Consider }}\left( {\frac{1}{b} + \frac{1}{c} - \frac{1}{a}} \right)\left( {\frac{1}{c} + \frac{1}{a} - \frac{1}{b}} \right) \cr & = \left( {\frac{1}{b} + \frac{2}{b} - \frac{1}{a} - \frac{1}{a}} \right)\left( {\frac{2}{b} - \frac{1}{b}} \right) \cr & {\text{Using }}\frac{1}{a} + \frac{1}{c} = \frac{2}{b} = \left( {\frac{3}{b} - \frac{2}{a}} \right)\left( {\frac{1}{b}} \right) = \frac{3}{{{b^2}}} - \frac{2}{{ab}} \cr} $$

If $$p, q, r$$  are in A.P. then in an A.P. or a G.P. or an H.P. $${a_1},{a_2},{a_3},.....,$$    etc., the terms $${a_p},{a_q},{a_r}$$  are in A.P., G.P. or H.P. respectively.

$$1^{st}$$ term $$→ 1,$$  $$2^{nd}$$ term $$= 2,$$  $$4^{th}$$ term $$→ 3,$$  $$7^{th}$$ term $$→ 4,$$  $$11^{th}$$ term $$→ 5, . . . . .$$
Series is $$1, 2, 4, 7, 11, . . . . .$$
$${a_n} = 1 + \frac{{n\left( {n - 1} \right)}}{2} = \frac{{{n^2} - n + 2}}{2}$$
If $$n = 14,$$  then $$a_n = 92,$$
If $$n = 15,$$  then $$a_n = 106.$$

$${1^3} = 1 \cdot \left( {1 - 1} \right) + 1,{2^3} = \left( {2 \cdot 1 + 1} \right) + 5,{3^3} = \left( {3 \cdot 2 + 1} \right) + 9 + 11,{4^3} = \left( {4 \cdot 3 + 1} \right) + 15 + 17 + 19,{\text{e}}{\text{.t}}{\text{.c}}{\text{.}}$$
$$\therefore \,\,{n^3} = \left\{ {n \cdot \left( {n - 1} \right) + 1} \right\} + .....,$$       next term being 2 more than the previous
$$\therefore \,\,{n^3} = \left( {{n^2} - n + 1} \right) + \left( {{n^2} - n + 3} \right) + .....$$

$$\eqalign{ & {\text{Here, }}B - 2A = \sum\limits_{n = 1}^{40} {{a_n} - 2\sum\limits_{n = 1}^{20} {{a_n}} = \sum\limits_{n = 21}^{40} {{a_n}} - 2\sum\limits_{n = 1}^{20} {{a_n}} } \cr & B - 2A = \left( {{{21}^2} + {{2.22}^2} + {{23}^2} + {{2.24}^2} + ..... + {{40}^2}} \right) - \left( {{1^2} + {{2.2}^2} + {3^2} + {{2.4}^2} + ..... + {{20}^2}} \right) \cr & = 20\left[ {22 + 2.24 + 26 + 2.28 + ..... + 60} \right] \cr & = 20\left[ {\underbrace {\left( {22 + 24 + 26 + ..... + 60} \right)}_{20{\text{ terms}}} + \underbrace {\left( {24 + 28 + ..... + 60} \right)}_{10{\text{ terms}}}} \right] \cr & 20\left[ {\frac{{20}}{2}\left( {22 + 60} \right) + \frac{{10}}{2}\left( {24 + 60} \right)} \right] \cr & = 10\left[ {20.82 + 10.84} \right] \cr & = 100\left[ {164 + 84} \right] \cr & = 100.\left( {248} \right) \cr} $$

Let two numbers be $$a$$ and $$b.$$
Given, $$\frac{{2ab}}{{a + b}} = 4$$
$$\eqalign{ & \Rightarrow ab = 2\left( {a + b} \right) \cr & 2A + {G^2} = 27 \cr & \Rightarrow 2\left( {\frac{{a + b}}{2}} \right) + ab = 27 \cr & \Rightarrow ab = 18{\text{ and }}a + b = 9 \cr & \Rightarrow ab = 9 \cr} $$
On solving these we get
$$a = 3\,\,\&\,\, b = 6\,\,{\text{or }}a = 6\,\,\&\,\, b = 3.$$