Solution:

2016 is a leap year.
∴ It has 366 days.

Now, Sanika’s daily savings are:
10, 11, 12, ... 366
which is an A.P.

Here,
\(t_1 = a = 10, d = 1,\)
\(n = 366 S_n =\ ?\)

Now, \(\displaystyle S_n = \frac {n}{2}[2a + (n - 1)d]\)

∴ \(\displaystyle S_n = \frac {\cancelto {183}{366}}{\cancelto {1}{2}}\:[2 \times 10 + (366 - 1)\times 1]\)

∴ \(S_n = 183\:[20 + 365]\)

∴ \(S_n = 183 \times 385\)

∴ \(S_n = 70455\)

∴ Sanika’s total savings on 31 Dec. 2016 would be ₹ 70,455/-

Solution:

Total amount repaid
= ₹ 8000 + ₹ 1360 = ₹ 9360

Here,
\(n = 12\), \(d = - 40\), \(S_n = 9360\),
\(t_1 =\ ?\), \(t_{12} =\ ?\),

Now,
 \(\displaystyle S_n = \frac {n}{2}[2a + (n - 1)d]\)

∴ \(\displaystyle S_{12} = \frac {\cancelto {6}{12}}{\cancelto {1}{2}}[2a + (12 - 1) \times - 40]\)

∴ \(9360 = 6\:[2a + 11 \times -40]\)
∴ \(9360 = 6\:[2a -\:440]\)
i.e. \(6\:[2a -\:440] = 9360\)

∴ \(\displaystyle 2a - 440 = \frac {\cancelto {1560}{9360}}{\cancelto {1}{6}}\)

∴ \(2a - 440 = 1560\)
∴ \(2a = 1560 + 440\)
∴ \(2a = 2000\)

∴ \(\displaystyle a = \frac {\cancelto {1000}{2000}}{\cancelto {1}{2}}\)

∴ \(a = t_1 = 1000\) ... (i)

Also, \(t_n = a + (n - 1)d\)
∴ \(t_{12} = 1000 + (12 - 1)\times -\:40\)
∴ \(t_{12} = 1000 + 11 \times -\:40\)
∴ \(t_{12} = 1000 -\:440\)
∴ \(t_{12} = 560\) ... (ii)

∴ The first instalment is ₹ 1000 and the last instalment is ₹ 560.

Solution:

This is an A. P. where,
\(t_1 = a = 5000\), \(t_2 = 7000\), \(t_3 = 9000\)
\(t_3 = 9000\), \(S_{12} =\ ?\)

Here,
\(d = t_2 - t_1 = 7000 - 5000 = 2000\)

Now,
 \(\displaystyle S_n = \frac {n}{2}[2a + (n - 1)d]\)

∴ \(\displaystyle S_{12} = \frac {12}{2}[2 \times 5000 + (12 - 1) \times 2000]\)

∴ \(\displaystyle S_{12} = \frac {\cancelto {6}{12}}{\cancelto {1}{2}} \times [10000 + 11 \times 2000]\)

∴ \(S_{12} = 6\:[10000 + 22000]\)

∴ \(S_{12} = 6 \times 32000\)

∴ \(S_{12} = 192000\)

∴ The total amount invested by Sachin in 12 years is ₹ 1,92,000/-.

Solution:

Here,
\(a = t_1 = 20, t_2 = 22, t_3 = 24,\)
\(n = 27, t_{15} = ?, S_{27} = ?\)

Now, \(d = t_2 - t_1 = 22 - 20 = 2\)

And, \(t_n = a + (n - 1)d\)
∴ \(t_{15} = 20 + (15 - 1)\times 2\)
∴ \(t_{15} = 20 + 14\times 2\)
∴ \(t_{15} = 20 + 28\)
∴ \(t_{15} = 48\) ... (i)

Also,
 \(\displaystyle S_n = \frac {n}{2}[2a + (n - 1)d]\)

∴ \(\displaystyle S_{27} = \frac {27}{2}[2 \times 20 + (27 - 1) \times 2]\)

∴ \(\displaystyle S_{27} = \frac {27}{2}[40 + 26 \times 2]\)

∴ \(\displaystyle S_{27} = \frac {27}{2} \times [40 + 52]\)

∴ \(\displaystyle S_{27} = \frac {27}{\cancelto {1}{2}}\times \cancelto {46}{92}\)

∴ \(S_{27} = 27 \times 46\)
∴ \(S_{27} = 1242\) ... (ii)

∴ There are 48 seats in the 15^th row and the total number of seats in the auditorium is 1242.

Solution:

Let, the temperatures from Monday through Saturday be as follows:

From the first condition,
 \(\bcancel {a} - d + \cancel {a + 4d} = \bcancel {a} + \cancel {a + 4d} + 5\)
∴ \(- d = 5\)
i.e. \(d = - 5\) ... (i)

From the second condition,
 \(a + d = - 30\) ... (ii)

Substututing the value of \(d\) in (ii),
 \(a + d = - 30\) ... (ii)
∴ \(a - 5 = - 30\)
∴ \(a = - 30 + 5\)
∴ \(a = - 25\) ... (iii)

The temperatures on the other days are:
Monday: \(a - d = - 25 - (- 5) = - 25 + 5 = - 20\)° C
Tuesday: \(a = - 25\)° C
Thursday: \(a + 2d = - 25 + 2 \times (- 5) = - 25 - 10 = - 35\)° C
Friday: \(a + 3d = - 25 + 3 \times (- 5) = - 25 - 15 = - 40\)° C
Saturday: \(a + 4d = - 25 + 4 \times (- 5) = - 25 - 20 = - 45\)° C

∴ The temperatures of Kargil from Monday through Saturday are − 20° C, − 25° C, − 30° C, − 35° C, − 40° C, − 45° C respectively.

Solution:

This is an A. P. like:
 1, 2, 3, ... , 23, 24, 25.

Here,
 \(t_1 = 1, t_2 = 2,\)
 \(t_{25} = 25, S_{25} = ?\)

Now,
 \(\displaystyle S_n = \frac {n}{2}[t_1 + t_n]\)

∴ \(\displaystyle S_{25} = \frac {25}{2}[1 + 25]\)

∴ \(\displaystyle S_{25} = \frac {25}{\cancelto {1}{2}} \times \cancelto {13}{26}\)

∴ \(S_{25} = 25 \times 13\)
∴ \(S_{25} = 325\)

∴ Total 325 trees were planted in 25 rows.