**Q.1: The sum of first q terms of an AP is ( 63q – \(3q^{ 2 }\) ) If its pth term is -60, find the value of p. Also, find the 11 ^{th} term of its AP.**

**Sol:**

Let \(S_{q}\)

Then,

\(S_{q} = 63q – 3q^{2} \)

\(\Rightarrow S_{q – 1} = 63 (q – 1) – 3(q – 1)^{2} \)

\( = 63q – 63 – 3(q^{2} – 2q + 1) \)

\( =-3q^{2} + 69q – 66 \)

Suppose \(a_{q}\)

Therefore, \(a_{q} = S_{q} – S_{q-1} \)

= \( 63q – 3q^{2} – (-3q^{2} + 69q – 66) \)

= – 6q + 66 ………………(i)

Now,

\(a_{p}\)

\(\Rightarrow -6p + 66 = – 60\)

\(\Rightarrow -6p = – 60 – 66 = – 126\)

\(\Rightarrow p = 21\)

Thus, the value of p is 21.

Putting q = 11 in (i), we get

\(a_{11} = -6 \times 11 + 66 = -66 + 66 = 0\)

Hence, the 11th term of the AP is 0.

**Q.2: Find the number of terms of the AP -12, -9, -6, …,21. If 1 is added to each term of this AP then find the sum of all terms of the AP thus obtained.**

**Sol:**

The given AP is -12, -9, -6,…….., 21.

Here, a = -12, d = -9 – (-12) = -9 + 12 = 3 and l = 21

Suppose there are n terms in the AP.

\(Therefore, l = a_{n} = 21\)

\(\Rightarrow -12 + (n – 1) \times 3 = 21\)

\(\Rightarrow 3n – 15 = 21\)

\(\Rightarrow 3n = 36 \)

\(\Rightarrow n = 12 \)

Thus, there are 12 terms in the AP.

If 1 is added to each term of the AP, then the new AP so obtained is -11, -8, -5, ……., 22.

Here, first term, A = -11, last term, L = 22 and n = 12

Therefore, Sum of the terms of this AP

\(= \frac{12}{2} (-11 + 22)\)

\(= 6 \times 11\)

= 66

Hence, the required sum is 66.

**Q.3: Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25 ^{th} term.**

**Sol:**

Let d be the common difference of the AP.

Here, a = 10 and n = 14

Now,

\(S_{14} = 1505\)

\(\Rightarrow \frac{14}{2} \left [ 2 \times 10 + (14 – 1) \times d \right ] = 1505\)

\(\Rightarrow 7 (20 + 13d) = 1505\)

\(\Rightarrow 20 + 13d = 215\)

\(\Rightarrow 13d = 195\)

\(\Rightarrow d = 15\)

Therefore, 25th term of the AP, \(a_{25}\)

= 10 + (25 – 1) \(\times\)

= 10 + 360

= 370

Hence, the required term is 370.

**Q.4: Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.**

**Sol:**

Let a be the first term and d be the common difference of the AP.

Then,

\(d = a_{3} – a_{2}\)

Now,

\(a_{2} = 14\)

\(\Rightarrow a + d = 14\)

\(\Rightarrow a + 4 = 14\)

\(\Rightarrow a = 10 \)

Using the formula, \(S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]\)

\(= \frac{51}{2} \left [ 2 \times 10 + (51 – 1) \times 4 \right ] \)

= \(= \frac{51}{2} \times (20 + 200) \)

\(= \frac{51}{2} \times 220 \)

\(= 5610 \)

Hence, the required sum is 5610.

**Q42 : In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by students. Which value is shown in the questions?**

**Sol:**

Number of trees planted by the students of each section of class 1 = 2

There are two sections of class 1.

Therefore, Number of trees planted by the students of class 1 = 2 \(\times\)

Number of trees planted by the students of each section of class 2 = 4

There are two sections of class 2.

Therefore, Number of trees planted by the students of class 2 = 2 \(\times\)

Similarly,

Number of trees planted by the students of class 3 = 2 \(\times\)

So, the number of trees planted by the students of different classes are 4, 8, 12, ……

Therefore, Total number of trees planted by the students = 4 + 8 + 12 + ….. up to 12 terms

This series is an arithmetic series.

Here, a = 4, d = 8 – 4 = 4 and n = 12

Using the formula, \(S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]\)

\(= \frac{12}{2} \left [ 2 \times 4 + (12 – 1) \times 4 \right ] \)

\(= 6 \times (8 + 44) \)

\(= 6 \times 52 \)

\(= 312 \)

Hence, the total number of trees planted by the students is 312.

**Q.5: In a potato race, a bucket is placed at the starting point, which is 5m from the first potato, and the other potatoes are placed 3m apart in a straight line. There are 10 potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and he continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?**

**Sol:**

Distance covered by the competitor to pick and drop the first potato = 2 \(\times\)

Distance covered by the competitor to pick and drop the second potato = 2 \(\times\)

Distance covered by the competitor to pick and drop the third potato = 2 \(\times\)

and so on.

Therefore, total distance covered by the competitor = 10m + 16m + 22m + …….. Up to 10 terms

This is an arithmetic series.

Here a = 10, d = 16 – 10 = 6 and n = 10

Using the formula, \(S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]\)

\(S_{10} = \frac{10}{2} \left [ 2 \times 10 + (10 – 1) \times 6 \right ] \)

\(= 5 \times (20 + 54) \)

\(= 5 \times 74 \)

\(= 370 \)

Hence, the total distance of the competitor has to run is 370 m.

**Q.6: There are 25 trees at equal distances of 5m in a line with a water tank, the distance of the water tank from the nearest tree being 10 m. A gardener waters all the trees separately,startingg from the water tank and returning back to the water tank after watering each tree to get water all the trees. **

**Sol:**

Distance covered by the gardener to water the first tree and return to the water tank = 10m + 10m = 20 m

Distance covered by the gardener to water the second tree and return to the water tank = 15m + 15 m = 30 m

Distance covered by the gardener to water the third tree and return to the water tank = 20m + 20m = 40 m

Therefore, total distance covered by the gardener to water all the trees = 20m + 30m + 40m + …….. Up to 25 terms

This is an arithmetic series.

Here a = 20, d = 30 – 20 = 10 and n = 25

Using the formula, \(S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]\)

\(S_{25} = \frac{25}{2} \left [ 2 \times 20 + (25 – 1) \times 10 \right ] \)

\(= \frac{25}{2} \times (40 + 240) \)

\(= \frac{25}{2} \times 280 \)

\(= 3500 \)

Hence, the total distance covered by the gardener to water all the trees is 3500 m.

**Q.7: A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each prize.**

**Sol:**

Let the value of the first prize be Rs. a

Since the value of each prize is Rs. 20 less than its preceding prize, so the values of the prizes are in AP with the common difference -Rs. 20

\(Therefore, d = -20\)

Number of cash prize to be given to the students, n = 7

Total sum of the prizes, \(S_{7} = Rs. 700\)

Using the formula, \(S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]\)

\(S_{7} = \frac{7}{2} \left [ 2 \times a + (7 – 1) \times (-20) \right ] = 700 \)

\(= \frac{7}{2} \times (2 a – 120) = 700 \)

\(= 7 a – 420 = 700 \)

\(\Rightarrow 7a = 700 + 420 = 1120\)

\(\Rightarrow a = 160 \)

Thus, the value of the first prize is Rs. 160.

Hence, the value of each prize is Rs. 160, Rs. 140, Rs. 120,Rs. 100, Rs. 80, Rs. 60, and Rs. 40.

**Q.8: A man saved 33000 in 10 months. In each month after the the first, he saved Rs. 100 more than he did in the 3 preceding month. How much did he save in the first month?**

**Sol:**

Let the money saved by the man in the first month be Rs. a.

It is given that in each month after the first, he saved Rs. 100 more than he did in the preceding month. So, the money saved by the man every month is in AP with common difference Rs. 100.

\(Therefore, d = 100\)

Number of month, n = 10

Total sum of money saved in 10 months, \(S_{10} = Rs. 33,000\)

Using the formula, \(S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]\)

\(S_{10} = \frac{10}{2} \left [ 2 \times a + (10 – 1) \times 100 \right ] = 33000 \)

\(5 \times (2 a + 900) = 33000 \)

\(\Rightarrow 2a + 900 = 6600 \)

\(\Rightarrow 2a = 6600 – 900 = 5700 \)

\(\Rightarrow a = 2850 \)

Hence, the money saved by the man in the first months is Rs. 2850.

**Q.9: A man arranges to pay off a debt of Rs. 36000 by 40 monthly instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid. Find the value of the first instalment.**

**Sol:**

Let the values of the first instalments be Rs. a.

Since the monthly instalments form an arithmetic series, so let us suppose the man increases the value of each instalment by Rs. d every month.

Therefore, Common difference of the arithmetic series = – Rs.d

Amount paid in 30 instalments = Rs. 26,000 – \(\frac{1}{3}\times 36000 – 12000\)

Let \(S_{n}\)

\(S_{30}\)

\(\Rightarrow \frac{30}{2} \left [ 2a + (30 – 1) d \right ] = 24000\)

\(\Rightarrow 15 \left [ 2a + 29d \right ] = 24000\)

\(\Rightarrow 2a + 29d = 1600\)

Also,

\(S_{40}\)

\(\Rightarrow \frac{40}{2} \left [ 2a + (40 – 1) d \right ] = 36000\)

\(\Rightarrow 20 \left [ 2a + 39d \right ] = 36000 \)

\(\Rightarrow 2a + 39d = 1800\)

Subtracting (i) from (ii), we get

\( (2a + 39d) – (2a + 29d) = 1800 – 1600 \)

\(\Rightarrow 10 d = 200\)

\(\Rightarrow d = 20 \)

Putting d = 20 in (i), we get

\(2a + 29 \times 20 = 1600\)

\(\Rightarrow 2a + 580 = 1600\)

\(\Rightarrow 2a = 1020\)

\(\Rightarrow a = 510 \)

Thus, the value of the first instalment is Rs. 510

**Q.10: A contract on construction job specifies a penalty for delay of completion beyond date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc. the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days? **

**Sol:**

It is given that the penalty for each succeeding day is Rs. 50 more than the preceding day, so the amount of penalties are in AP with common difference Rs. 50

Number of days in the delay of the work = 30

The amount of penalties are Rs. 200, Rs. 250, Rs. 300, Rs. 350, ….. Up to 30 terms.

Therefore, total amount of money paid by the contractor as penalty, \(S_{30}\)

Here, a = Rs. 200, d = Rs. 50 and n = 30

Using the formula, \(S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]\)

\(S_{30} = \frac{30}{2} \left [ 2 \times 200 + (30 – 1) \times 50 \right ] \)

\(= 15 \times (400 + 1450) \)

\(= 15 \times 1850 \)

Hence, the contractor has to pay Rs. 27,750 as penatly.