First term of an A.P. (a) = 1
Last term (l) = 11
and sum of its terms = 36
Let n be number of terms and d be common difference, n
$$\eqalign{ & {a_n} = 1 = a + \left( {n - 1} \right)d = 11 \cr & \Rightarrow 1 + \left( {n - 1} \right)d = 11 \cr & \Rightarrow \left( {n - 1} \right)d = 11 - 1 \cr & \Rightarrow \left( {n - 1} \right)d = 10\,.....\,\left( 1 \right) \cr & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = 36 \cr} $$
$$ \Rightarrow \frac{n}{2}\left[ {2 \times 1 + 10} \right] = 36\,$$       $$\left[ {{\text{From}}\,\left( 1 \right)} \right]$$
$$\eqalign{ & \Rightarrow n\left( {2 + 10} \right) = 72 \cr & \Rightarrow 12n = 72 \cr & \Rightarrow n = \frac{{72}}{{12}} \cr & \Rightarrow n = 6 \cr} $$

The A.P. is 3, 7, 11, 15, ......
Where a = 3, d = 7 - 3 = 4 and sum S_n = 406
$$\eqalign{ & \refore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & \Rightarrow 406 = \frac{n}{2}\left[ {2 \times 3 \times \left( {n - 1} \right) \times 4} \right] \cr & \Rightarrow 812 = n\left( {6 + 4n - 4} \right) \cr & \Rightarrow 812 = n\left( {4n + 2} \right) \cr & \Rightarrow 4{n^2} + 2n - 812 = 0 \cr & \Rightarrow 2{n^2} + n - 406 = 0 \cr & \Rightarrow 2{n^2} + 29n - 28n - 406 = 0 \cr & \Rightarrow n\left( {2n + 29} \right) - 14\left( {2n + 29} \right) = 0 \cr & \Rightarrow \left( {2n + 29} \right)\left( {n - 14} \right) = 0 \cr & \refore n = 14\,{\text{or}}\,\frac{{ - 29}}{2} \cr & {\text{But}}\,n = \frac{{ - 29}}{2}\,{\text{is}}\,{\text{not}}\,{\text{possible}} \cr} $$

$$\eqalign{ & {\text{Sum}}\,{\text{of}}\,n\,{\text{terms}}\left( {{S_n}} \right) = 3{n^2} + 5n \cr & \refore {\text{Sum}}\,{\text{of}}\,\left( {n - 1} \right)\,{\text{terms}}\,\left( {{S_{n - 1}}} \right) \cr & = 3{\left( {n - 1} \right)^2} + 5\left( {n - 1} \right) \cr & = 3\left( {{n^2} - 2n + 1} \right) + 5n - 5 \cr & = 3{n^2} - 6n + 3 + 5n - 5 \cr & = 3{n^2} - n - 2 \cr & \refore {n^{th}}\,{\text{term}} = {S_n} - {S_{n - 1}} \cr & \Rightarrow {a_n} = 3{n^2} + 5n - 3{n^2} + n + 2 \cr & \,\,\,\,\,\,\,\,\,\,{a_n} = 6n + 2,\,{\text{But}}\,{a_n} = 164 \cr & \Rightarrow 6n + 2 = 164 \cr & \Rightarrow 6n = 164 - 2 \cr & \Rightarrow 6n = 162 \cr & \refore n = \frac{{162}}{6} = {27^{th}}\,{\text{term}} \cr} $$

B is in G.P. with a = 2⁰, r = 2, n = 65
$$\eqalign{ & \refore {S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{2^0}\left( {{2^{65}} - 1} \right)}}{{2 - 1}} \cr & \refore B = {2^{65}} - 1 \cr & \Rightarrow B = A - 1 \cr & \refore A\,{\text{is}}\,{\text{larger}}\,{\text{n}}\,B\,{\text{by}}\,1 \cr} $$

Let cost of an equipment is Rs. 100
Now percentages of depreciation at 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 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, value of equipment at end of 10 years = 100 - 82.5 = 17.5
The total cost being
$$\eqalign{ & = {\text{Rs}}{\text{.}}\,\frac{{6,00,000}}{{100}} \times 17.5 \cr & = {\text{Rs}}{\text{.}}\,1,05,000 \cr} $$

$$\eqalign{ & {a_n} = a + \left( {n - 1} \right)d \cr & {a_9} = 449 \cr & \,\,\,\,\,\, = a + \left( {9 - 1} \right)d \cr & \,\,\,\,\,\, = a + 8d\,.....\left( 1 \right) \cr & {a_{449}} = 9 \cr & \,\,\,\,\,\,\,\,\, = a + \left( {449 - 1} \right)d \cr & \,\,\,\,\,\,\,\,\, = a + 448d\,.....\left( 2 \right) \cr & {\text{Subtracting}} \cr & 440d = - 440 \cr & \Rightarrow d = \frac{{ - 440}}{{440}} = - 1 \cr & {\text{and}}\,a + 8d = 449 \cr & \Rightarrow a \times 8 \times \left( { - 1} \right) = 449 \cr & \Rightarrow a = 449 + 8 = 457 \cr & \refore 0 = a + \left( {n - 1} \right)d \cr & \Rightarrow 0 = 457 + \left( {n - 1} \right)\left( { - 1} \right) \cr & \Rightarrow 0 = 457 - n + 1 \cr & \Rightarrow n = 458 \cr & \refore {458^{{\text{th}}}}\,{\text{term}} = 0 \cr} $$

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, n
$$\eqalign{ & \frac{{{S_n}}}{{{S_n}}} = \frac{{\frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]}}{{\frac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right]}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} \cr} $$
$$\refore \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} = \frac{{3n + 5}}{{5n + 7}}$$
$${\text{Substituting n = 2n - 1, n}}$$
$$\frac{{2{a_1} + \left( {2n - 2} \right){d_1}}}{{2{a_2} + \left( {2n - 2} \right){d_2}}} = $$     $$\frac{{3\left( {2n - 1} \right) + 5}}{{5\left( {2n - 1} \right) + 7}}$$
$$ \Rightarrow \frac{{{a_1} + \left( {n - 1} \right){d_1}}}{{{a_2} + \left( {n - 1} \right){d_2}}} = $$     $$\frac{{6n - 3 + 5}}{{10n - 5 + 7}}$$    (Dividing by 2)
$$\eqalign{ & \Rightarrow \frac{{{a_{1n}}}}{{{a_{2n}}}} = \frac{{6n + 2}}{{10n + 2}} \cr & \Rightarrow \frac{{{a_{1n}}}}{{{a_{2n}}}} = \frac{{3n + 1}}{{5n + 1}} \cr} $$

$$\eqalign{ & {5^{th}}\,{\text{term}}\,{\text{of}}\,{\text{GP}} \cr & = a{r^{5 - 1}} \cr & = 16 \times {r^4} \cr & = 81 \cr & {\text{Or}},\,r = {\left( {\frac{{81}}{{16}}} \right)^{\frac{1}{4}}} = \frac{3}{2} \cr & {4^{th}}\,{\text{term}}\,{\text{of}}\,{\text{GP}} \cr & = a{r^{4 - 1}} \cr & = 16 \times {\left( {\frac{3}{2}} \right)^3} \cr & = 54 \cr} $$

Let first term of AP be a and common difference = d
7^th term = A₇ = a + 6d
11^th term = A₁₁ = a + 10d
According to question,
⇒ 7 × (a + 6d) = 11 × (a + 10d)
⇒ 7a + 42d = 11a + 110d
⇒ 11a - 7a = 42d - 110d
⇒ 4a = -68d
⇒ a = -17d
⇒ a + 17d = 0 = A₁₈

$$\eqalign{ & {S_n} = \frac{n}{2}\left[ {a + l} \right] \cr & {S_{13}} = \frac{{13}}{2}\left[ { - 10 + 26} \right] \cr & {S_{13}} = \frac{{13}}{2}\left[ {16} \right] \cr & {S_{13}} = 104 \cr} $$