NCERT Solutions for Maths Class 10 Chapter 5

Exercise 5.1

Question 1: In which of the following situations, does the list of numbers involved make as arithmetic progression and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\)  of the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Answer: 

(i) It can be observed that

Taxi fare for first km = 15

Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39

Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.

(ii) Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes  \(\frac{1}{4}\)  of air remaining in the cylinder at a time.

In other words, after every stroke, only \(1-\frac{1}{4}=\frac{3}{4}\) th part of air will remain.

Therefore, volumes will be  \(V, (\frac{3V}{4}), (\frac{3}{4} V)^2, (\frac{3}{4}V)^3\)….

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P. 

(iii) Cost of digging for first metre = 150

Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300

Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

(iv) We know that if Rs \(P\) is deposited at \(r%\) compound interest per annum for \(n\) year, our money will be \(P (1+\frac{r}{100p})^n\)  after n years.

Therefore, after every year, our money will be

\(10000(1+\frac{8}{100}), 10000(1+\frac{8}{100})^2, 10000(1+\frac{8}{100})^3, 10000(1+\frac{8}{100})^4……\)

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

Question 2: Write first four terms of the A.P. when the first term  \(a\) and the common difference d are given as follows

(i) \(a = 10,\space d = 10\)
(ii) \(a = − 2,\space d = 0\)
(iii) \(a = 4,\space d = − 3\)

(iv) \(a = − 1,\space d =\frac{1}{2}\)

(v) \(a = − 1.25,\space d = − 0.25\)

Answer:

(i) \(a = 10,\space d = 10\)

Let the series be \(a_1, a_2, a_3, a_4, a_5 … a_1 = a = 10\)

\(a_2 = a_1 + d = 10 + 10 = 20\)
\(a_3 = a_2 + d = 20 + 10 = 30\)
\(a_4 = a_3 + d = 30 + 10 = 40\)
\(a_5 = a_4 + d = 40 + 10 = 50\)

Therefore, the series will be \(10, 20, 30, 40, 50 …\)

First four terms of this A.P. will be \(10, 20, 30\) and \(40\).

(ii) \(a = −2,\space d = 0\)

Let the series be \(a_1, a_2, a_3, a_4 …\)

\(a_1 = a = −2\)

\(a_2 = a_1 + d = − 2 + 0 = −2\)
\(a_3 = a_2 + d = − 2 + 0 = −2\)
\(a_4 = a_3 + d = − 2 + 0 = −2\)

Therefore, the series will be \(−2, −2, −2, −2 …\)

First four terms of this A.P. will be \(−2, −2, −2\) and \(−2\).

(iii) \(a = 4,\space d = −3\)

Let the series be \(a_1, a_2, a_3, a_4 …\)
\(a_1 = a = 4\)

\(a_2 = a_1 + d = 4 − 3 = 1\)
\(a_3 = a_2 + d = 1 − 3 = −2\)
\(a_4 = a_3 + d = − 2 − 3 = −5\)

Therefore, the series will be \(4, 1, −2 −5 …\)

First four terms of this A.P. will be \(4, 1, −2\) and \(−5\).

(iv) \(a=-1,\space d=\frac{1}{2}\)

Let the series be \(a_1,a_2,a_3,a_4….\)

\(a_1=a=-1\)

\(a_2=a_1+d=-1+\frac{1}{2}=-\frac{1}{2}\)

\(a_3=a_2+d=\frac{1}{2}+\frac{1}{2}=0\)

\(a_4=a_3+d=0+\frac{1}{2}=\frac{1}{2}\)

Clearly, the series will be

\(-1, -\frac{1}{2}, 0, \frac{1}{2}………….\)

First four terms of this A.P. will be \(-1, -\frac{1}{2}, 0\) and \(\frac{1}{2}\)

(v) \(a = −1.25,\space d = −0.25\)

Let the series be \(a_1, a_2, a_3, a_4 …\)

\(a_1 = a = −1.25\)

\(a_2 = a_1 + d = − 1.25 − 0.25 = −1.50\)
\(a_3 = a_2 + d = − 1.50 − 0.25 = −1.75\)

\(a_4 = a_3 + d = − 1.75 − 0.25 = −2.00\)

Clearly, the series will be

\(1.25, −1.50, −1.75, −2.00 ……..\)

First four terms of this A.P. will be \(−1.25, −1.50, −1.75\) and \(−2.00\).

Question 3: For the following A.P.s, write the first term and the common difference.

(i) \(3,\space1, \space-1,\space-3….\)

(ii) \(-5,\space-1,\space3,\space7….\)

(iii) \(\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3}….\)

(iv) \(0.6,\space1.7,\space2.8,\space3.9….\)

Answer:

(i) \(3, 1, −1, −3 …\)

Here, first term, \(a = 3\)

Common difference, \(d =\) Second term − First term \(= 1 − 3 = −2\)

(ii) \(−5, −1, 3, 7 …\)

Here, first term, \(a = −5\)

Common difference, \(d =\) Second term – First term \(=(−1)−(−5)=−1+5=4\)

(iii) \(\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3}…..\)

Here, first term, \(a=\frac{1}{3}\)

Common difference, \(d=\) Second term – First term

(iv) \(0.6,  1.7,  2.8,  3.9  …\)
Here, first term, \(a = 0.6\)

Common difference, \(d =\) Second term − First term \(=1.7−0.6=1.1\)

Exercise 5.2

Question 1: Fill in the blanks in the following table, given that \(a\) is the first term, \(d\) the common difference and an the \(n^{th}\)  term of the A.P.

Answer: 

I. \(a = 7, d = 3, n = 8, an = ?\)

We know that,

For an A.P. \(an= a + (n − 1) d\)

\(= 7 + (8 − 1) 3\)

\(= 7 + (7) 3\)

\(= 7  +  21  =  28\)
Hence, \(an = 28\)

Correct: \(an=28\)

II. Given that

\(a = −18, n = 10, an = 0, d = ?\)

We know that,

\(an = a + (n − 1) d\)

\(0 = − 18 + (10 − 1) d\)

\(18 = 9d\)

Hence, common difference, \(d = 2\)

Correct: \(d=2\)

III.       Given that

\(d = −3, n = 18, an = −5\)

We know that,

\(an = a + (n − 1) d\)

\(−5 = a + (18 − 1) (−3)\)
\(−5 = a + (17) (−3)\)
\(−5 = a − 51\)

\(a = 51 − 5 = 46\)
Hence, \(a = 46\)

Correct: \(a=46\)

IV. \(a = −18.9, d = 2.5, an = 3.6, n = ?\)

We know that,

\(an = a + (n − 1) d\)

\(3.6 = − 18.9 + (n − 1) 2.5\)

\(3.6 + 18.9 = (n − 1) 2.5\)

\(22.5 = (n − 1) 2.5\)

\((n-1)=\frac{22.5}{2.5}\)

Hence,

Correct: \(n=10\)

V. \(a = 3.5, d = 0, n = 105, an = ?\)

We know that,

\(an = a + (n − 1) d\)

\(an = 3.5 + (105 − 1) 0\)
\(an = 3.5 + 104 \times 0\)
an = 3.5

Hence, \(an = 3.5\)

Correct: \(an=3.5\)

Question 2: Choose the correct choice in the following and justify

(i) \(30^{th}\) tern of the A.P: \(10, 7, 4,\) …… is

(a) \(97\)   (b) \(77\)      (c) \(-77\)          (d) \(87\)

(ii) \(11^{th}\) term of th A.P.

\(-3,-\frac{1}{2},2,……..\) is

(a) \(28\)     (b) \(22\)   (c) \(-38\)    (d) \(-8\frac{1}{2}\)

Answer: 

Give that

A.P. \(10,\space7,\space4,……\)

First term, \(a=10\)

Common difference, \(d=a_2-a_1=7-10=-3\)

We know that, \(an=a+(n-1)d\)

\(a_{30} = 10 + (30 − 1) (−3)\)
\(a_{30} = 10 + (29) (−3)\)
\(a_{30} = 10 − 87 = −77\)

Hence, the correct answer is C.

(ii) Given taht, A.P. \(-3,-\frac{1}{2},2\)

First term \(a=-3\)

Common difference, \(d=a_2-a_1\)

\(=-\frac{1}{2}-(-3)\)

\(=-\frac{1}{2}+3=\frac{5}{2}\)

We know that, 

\(a_n=a+(n-1)d\)

\(a_{11}=-3+(11-1)(\frac{5}{2})\)

\(a_{11}=-3+(10)(\frac{5}{2})\)

\(a_{11}=-3+25\)

\(a_{11}=22\)

Hence, the answer is B.

Question 3: In the following APs find the missing term in the boxes.

(i) \(2,\space□,\space26\)

(ii) \(□,\space13,□,\space,3\)

(iii) \(5,\space□,\space□,9\frac{1}{2}\)

(iv) \(-4,\space□,\space□,\space□,\space□,\space6\) 

(v) \(□,\space28,\space,□,□,□,\space-22\)

Answer: 

(i) \(2, □, 26\)

For this A.P.,

a = 2

a3 = 26

We know that, an = a + (n − 1) d
a3 = 2 + (3 − 1) d

26 = 2 + 2d

24 = 2d

d = 12

a2 = 2 + (2 − 1) 12
= 14

Therefore, 14 is the missing term.

(ii) \(□, 13, □,3\)

For this A.P.,
\(a_2 = 13\) and \(a_4 = 3\)

We know that, \(an = a + (n − 1) d\)
\(a_2 = a + (2 − 1) d\)

\(13 = a + d\)      ….(i)

\(a_4 = a + (4 − 1) d\)
\(3 = a + 3d\)            ….(ii)

On subtracting (i) from (ii), we obtain

\(−10 = 2d\)

\(d = −5\)

From equation (i), we obtain

\(13 = a + (−5)\)

\(a = 18\)

\(a_3 = 18 + (3 − 1) (−5)\)

\(= 18 + 2 (−5) = 18 − 10 = 8\)

Therefore, the missing terms are 18 and 8 respectively.

(iii) \(5,□, □, 9\frac{1}{2}\)

For the A.P.

\(a=5\)

\(a_4=0\frac{1}{2}=\frac{19}{2}\)

We know that,

\(a_n=a+n(n-1)d\)

\(a_4=a+(4+1)d\)

\(\frac{19}{2}=5+3d\)

\(\frac{19}{2}-5=3d\)

\(\frac{9}{2}=3d\)\

\(d=\frac{3}{2}\)

\(a_2=a+d=5+\frac{3}{2}=\frac{13}{2}\)

\(a_3=a+2d=5+2(\frac{3}{2})=8\)

Therefore, the missing terms are \(\frac{13}{2}\) and \(8\) respectively.

(iv) \(-4,□, □, □, □, 6\)

For this A.P.,

\(a = −4\) and \(a_6 = 6\)

We know that,

\(an = a + (n − 1) d\)
\(a_6 = a + (6 − 1) d\)
\(6 = − 4 + 5d\)

\(10 = 5d\)

\(d = 2\)

\(a_2 = a + d = − 4 + 2 = −2\)

\(a_3 = a + 2d = − 4 + 2 (2) = 0\)
\(a_4 = a + 3d = − 4 + 3 (2) = 2\)
\(a_5 = a + 4d = − 4 + 4 (2) = 4\)

Therefore, the missing terms are \(−2, 0, 2\) and \(4\) respectively.

(v) \(□,38, □, □, □, -22\)

For the A.P.,

\(a_6 = −22\)

We know that

\(an = a + (n − 1) d\)
\(a_2 = a + (2 − 1) d\)
\(38 = a + d\)       …..(i)
\(a_6 = a + (6 − 1) d\)
\(−22 = a + 5d\)        …..(ii) 

On subtracting equation (i) from (ii), we obtain \(− 22 − 38 = 4d\)

\(−60 = 4d\)

\(d = −15|)

\(a = a_2 − d = 38 − (−15) = 53\)

\(a_3 = a + 2d = 53 + 2 (−15) = 23\)
\(a_4 = a + 3d = 53 + 3 (−15) = 8\)
\(a_5 = a + 4d = 53 + 4 (−15) = −7\)

Therefore, the missing terms are \(53, 23, 8\) and \(−7\) respectively.

Question 4: Which term of the A.P. \(3, 8, 13, 18,\) … is \(78\)?

Answer:

\(3, 8, 13, 18, …\)
For this A.P.,
\(a = 3\)

\(d = a_2 − a1 = 8 − 3 = 5\)

Let n^th term of this A.P. be \(78\).
\(an = a + (n − 1) d\)

\(78 = 3 + (n − 1) 5\)
\(75 = (n − 1) 5\)
\((n − 1) = 15\)
\(n = 16\)

Hence, 16^th term of this A.P. is \(78\).

Question 5:  Find the number of terms in each of the following A.P.

(i) \(7,13,19,…..,205\)

(ii) \(18,15\frac{1}{2},13,……,-47\)

Answer: 

(i) \(7, 13, 19, …, 205\)

For this A.P.,

\(a = 7\)

\(d = a_2 − a_1 = 13 − 7 = 6\)

Let there are n terms in this A.P.

\(an = 205\)

We know that

\(an = a + (n − 1) d\)

Therefore, \(205 = 7 + (n − 1) 6\)
\(198 = (n − 1) 6\)

\(33 = (n − 1)\)
\(n = 34\)

Therefore, this given series has \(34\) terms in it.

(ii) \(18,15\frac{1}{2},13……,-47\)

For this A.P,

\(a=18\)

\(d=a_2-a_1=15\frac{1}{2}-18\)

\(d=\frac{31-36}{2}=-\frac{5}{2}\)

Let there are \(n\) terms in this A.P.

Therefore, \(an=-47\) and we know that,

\(a_n=a+(n-1)d\)

\(-47=18+(n-1)(-\frac{5}{2})\)

\(-47-1(n-1)(-\frac{5}{2})\)

\(-65=(n-1)(-\frac{5}{2})\)

\((n-1)=\frac{-136}{-5}\)

\((n-1)=26\)

\(n=27\)

Therefore, this given A.P. has 27 terms in it.

Question 6: Check whether \(− 150\) is a term of the A.P. \(11, 8, 5, 2, …\)

For this A.P 

\(a=11\)

\(d=a_2-a_1=88-11=-3\)

Let \(-150\) be th n^th terms of this A.P.,

We know that,

\(a_n=a+(n-1)d\)

\(-150=11+(n-1)(-3)\)

\(-150=11-3n+3\)

\(-164==-3n\)

\(n=\frac{164}{3}\)

Clearly, \(n\) is not an integer.

Therefore, \(-150\) is not a term of this A.P.

Question 7: Find the 3^1st term of an A.P. whose 11^th term is 38 and the 16^th term is 73

Answer:

Given that,

\(a_{11} = 38\)

\(a_{16} = 73\)

We know that,

\(an = a + (n − 1) d\)

\(a_{11} = a + (11 − 1) d\)
\(38 = a + 10d\)       ….(i)

Similarly,

\(a_{16} = a + (16 − 1) d\)
\(73 = a + 15d\)         …….(ii)

On subtracting (i) from (ii), we obtain \(35 = 5d\)

\(d = 7\)

From equation (i),

\(38 = a + 10 × (7)\)

\(38 − 70 = a\)

\(a = −32\)

\(a_{31} = a + (31 − 1) d\)
\(= − 32 + 30 (7)\)
\(= − 32 + 210\)

\(= 178\)

Hence,  term is \(178\).

Question 8: An A.P. consists of 50 terms of which 3^rd term is 12 and the last term is 106.

Find the 29^th term.

Answer:

Given that,

\(a_3 = 12\)

\(a_{50} = 106\)

We know that,

\(an = a + (n − 1) d\)
\(a_3 = a + (3 − 1) d\)
\(12 = a + 2d\)            ..(i)

Similarly, \(a_{50} = a + (50 − 1) d\)

\(106 = a + 49d\)           …(ii)

On subtracting (i) from (ii), we obtain \(94 = 47d\)

\(d = 2\)

From equation (i), we obtain

\(12 = a + 2 (2)\)

\(a = 12 − 4 = 8\)

\(a_{29} = a + (29 − 1) d\)
\(a_{29} = 8 + (28)2\)
\(a_{29} = 8 + 56 = 64\)

Therefore, 29^th term is \(64\).

Question 9:

If the 3^rd and the 9^th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.

Answer:

Given that,
\(a_3 = 4\)

\(a_9 = −8\)

We know that,

\(an = a + (n − 1) d\)
\(a_3 = a + (3 − 1) d\)
\(4 = a + 2d\)        ….(i)
\(a_9 = a + (9 − 1) d\)
\(−8 = a + 8d\)          …(ii)

On subtracting equation (i) from (ii), we obtain \(−12 = 6d\)

\(d = −2\)

From equation (i), we obtain
\(4 = a + 2 (−2)\)

\(4 = a − 4\)

\(a = 8\)

Let n^th term of this A.P. be zero.
\(an = a + (n − 1) d\)

\(0 = 8 + (n − 1) (−2)\)
0 = 8 − 2n + 2\)

\(2n = 10\)

\(n = 5\)

Hence, 5^th term of this A.P. is 0.

Question 10:

If 17^th term of an A.P. exceeds its 10^th term by 7. Find the common difference.

Answer:

We know that,

For an A.P., \(an = a + (n − 1) d\)
\(a_{17} = a + (17 − 1) d\)

\(a_{17} = a + 16d\)

Similarly, \(a_{10} = a + 9d\)
It is given that

\(a_{17} − a10 = 7\)

\((a + 16d) − (a + 9d) = 7\)
\(7d = 7\)

\(d = 1\)

Therefore, the common difference is \(1\).

Question 11: Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54^th term?

Answer: Given A.P. is \(3, 15, 27, 39, …\)

\(a = 3\)

\(d = a_2 − a_1 = 15 − 3 = 12\)
\(a{54} = a + (54 − 1) d\)

\(= 3 + (53) (12)\)
\(= 3 + 636 = 639\)
\(132 + 639 = 771\)

We have to find the term of this A.P. which is \(771\). Let n^th term be \(771\).

\(an = a + (n − 1) d\)

\(771 = 3 + (n − 1) 12\)
\(768 = (n − 1) 12\)
\((n − 1) = 64\)

\(n = 65\)

Therefore, 65^th term was 132 more than 54^th term. Alternatively,

Let n^th term be 132 more than 54^th term.

\(n=54+\frac{132}{12}\)

\(=54+11\)

\(=65^{th}\) term

Question 12: Two A.P.s have the same common difference. The difference between their 100^th term is 100, what is the difference between their 1000^th terms?

Answer:  Let the first term of these A.P.s be a1 and \(a_2\) respectively and the common difference of these A.P.s be \(d\).

For first A.P.,

\(a_{100} = a1 + (100 − 1) d\)

\(= a_1 + 99d\)

\(a_{1000} = a_1 + (1000 − 1) d\)
\(a_{1000} = a_1 + 999d\)

For second A.P.,

\(a_{100} = a_2 + (100 − 1) d\)
\(= a_2 + 99d\)

\(a_{1000} = a_2 + (1000 − 1) d\)
\(= a_2 + 999d\)

Given that, difference between

100^th term of these A.P.s \(= 100\)

Therefore, \((a_1 + 99d) − (a_2 + 99d) = 100 a_1 − a_2 = 100\)  ..(i)

Difference between 1000^th terms of these A.P.s \((a_1+999d)−(a_2+999d)=a_1−a_2\)
From equation (i),

This difference, \(a_1 − a_2 = 100\)

Hence, the difference between 1000^th terms of these A.P. will be 100.

Question 13: How many three digit numbers are divisible by 7

Answer:

First three-digit number that is divisible by \(7=105\)

Next number \(=105+7=112\)

Therefore, \(105, 112, 119, …\)

All are three digit numbers which are divisible by 7 and thus, all these
are terms of an A.P. having first term as  105 and common difference as 7.

The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5.

Clearly, \(999 − 5 = 994\) is the maximum possible three-digit number that is divisible by 7.

The series is as follows.

\(105, 112, 119, …, 994\)

Let 994 be the nth term of this A.P.

\(a = 105\)

\(d = 7\)

\(an = 994\)
\(n = ?\)

\(an = a + (n − 1) d\)

\(994 = 105 + (n − 1) 7\)
\(889 = (n − 1) 7\)

\((n − 1) = 127\) 
\(n = 128\)

Therefore, 128 three-digit numbers are divisible by 7.

Question 14:

How many multiples of 4 lie between 10 and 250?

Answer: First multiple of \(4\) that is greater than \(10\) is \(12\). Next will be \(16\).

Therefore, \(12, 16, 20, 24, …\)

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.

When we divide \(250\) by \(4\), the remainder will be \(2\).

Therefore, \(250 − 2 = 248\) is divisible by \(4\).

The series is as follows.

\(12, 16, 20, 24, …, 248\)

Let 248 be the n^th term of this A.P.

\(a=12\)

\(d=4\)

\(a_n=248\)

\(a_n=a+(n-1)d\)

\(248=12+(n-1)4\)

\(\frac{236}{4}=n-1\)

\(59=n-1\)

\(n=60\)

Therefore, there are 60 multiples of 4 between 10 and 250.

Question 15: For what value of \(n\), are the n^th terms of two APs \(63, 65, 67\) and \(3, 10, 17, …\) equal

Answer:

\(63, 65, 67, …\)
\(a = 63\)

\(d = a_2 − a_1 = 65 − 63 = 2\)

n^th term of this A.P. \(= an = a + (n − 1)\)
\(d an= 63 + (n − 1) 2 = 63 + 2n − 2\)

\(an = 61 + 2n \)       ….(i)
\(3, 10, 17, …\)

\(a = 3\)

\(d = a2 − a1 = 10 − 3 = 7\)

n^th term of this A.P. \(= 3 + (n − 1) 7 an = 3 + 7n − 7\)

\(an = 7n − 4 (2)\)

It is given that, n^th term of these A.P.s are equal to each other.
Equating both these equations, we obtain

\(61 + 2n = 7n − 4\)

\(61 + 4 = 5n\)

\(5n = 65\)

\(n = 13\)

Therefore, 13^th terms of both these A.P.s are equal to each other.

 Question 16:

Determine the A.P. whose third term is 16 and the 7^th term exceeds the 5^th term by 12.

Answer: 

\(a_3 = 16\)

\(a + (3 − 1) d = 16\)
\(a + 2d = 16 \)      …..(i)

\(a_7 − a_5 = 12\)

\([a+ (7 − 1) d] − [a + (5 − 1) d]\)

\(= 12 (a + 6d) − (a + 4d) = 12\)

\(2d = 12\)

\(d = 6\)

From equation (i), we obtain

\(a + 2 (6) = 16\)

\(a + 12 = 16\)
\(a = 4\)

Therefore, A.P. will be \(4, 10, 16, 22, …\)

Question 17: Find the 20^th term from the last term of the A.P. \(3, 8, 13, …, 253\)

Answer:  Given A.P. is \(3, 8, 13, …, 253\)

Common difference for this A.P. is \(5\).

Therefore, this A.P. can be written in reverse order as \(253, 248, 243, …, 13, 8, 5\)

For this A.P.,

\(a = 253\)

\(d = 248 − 253 = −5\)
\(n = 20\)

\(a_{20} = a + (20 − 1) d\)
\(a_{20} = 253 + (19) (−5)\)
\(a_{20} = 253 − 95\)

\(a = 158\)

Therefore, 20^th term from the last term is 158.

Question 18: The sum of 4^th and 8^th terms of an A.P. is 24 and the sum of the 6th and 10^th terms is 44. Find the first three terms of the A.P.

Answer: We know that,

\(an = a + (n − 1) d\)
\(a_4 = a + (4 − 1) d\)
\(a_4 = a + 3d\)

Similarly,

\(a_8 = a + 7d\)
\(a_6 = a + 5d\)
\(a_{10} = a + 9d\)

Given that, \(a_4 + a_8 = 24\)

\(a + 3d + a + 7d = 24\)
\(2_a + 10d = 24\)

\(a + 5d = 12\)    ….(i)
\(a_6 + a_{10} = 44\)

\(a + 5d + a + 9d = 44\)
\(2_a + 14d = 44\)

\(a + 7d = 22\)           …(ii)

On subtracting equation (i) from (i), we obtain

\(2d = 22 − 12\)

\(2d = 10\)

\(d = 5\)

From equation (i), we obtain

\(a + 5d = 12\)

\(a + 5 (5) = 12\)
\(a + 25 = 12\)
\(a = −13\)

\(a_2 = a + d = − 13 + 5 = −8\)
\(a_3 = a_2 + d = − 8 + 5 = −3\)

Therefore, the first three terms of this A.P. are \(−13, −8\) and \(−3\).

Question 19:

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Answer: It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.

Therefore, the salaries of each year after 1995 are 5000, 5200, 5400, …

Here, \(a = 5000\)

\(d = 200\)

Let after n^th year, his salary be Rs 7000.

Therefore, \(an = a + (n − 1) d\)

\(7000 = 5000 + (n − 1) 200\)
\(200(n − 1) = 2000\)

\((n − 1) = 10\)

\(n = 11\)

Therefore, in 11th year, his salary will be Rs. \(7000\).

Question 20: Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the n^th week, her week, her weekly savings become Rs 20.75, find \(n\).

Answer:

Given that,

\(a = 5\)

\(d = 1.75\)

\(an = 20.75\)
\(n = ?\)

\(an = a + (n − 1) d\)

\(20.75=5+(n-1)1.75\)

\(15.75=(n-1)1.75\)

\((n-1)=\frac{15.75}{1.75}=\frac{1575}{175}\)

\(=\frac{63}{7}\)

\(n − 1 = 9\)

\(n = 10\)

Hence, \(n\) is \(10\).

Question 1: Find the sum of the following APs.

(i) \(2, 7, 12 ,….,\) to \(10\) terms.

(ii) \(- 37, – 33, – 29 ,…,\) to \(12\) terms

(iii) \(6, 1.7, 2.8 ,……..,\) to \(100\) terms

(iv) \(\frac{1}{15}, \frac{1}{12},\frac{1}{10},………,\) to \(11\) terms

Answer:

(i) \(2, 7, 12 ,…,\) to \(10\) terms

For this A.P.,

\(a = 2\)

\(d = a_2 − a_1 = 7 − 2 = 5\)
\(n = 10\)

We know that,

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

\(S_{10}=\frac{10}{2}[2(2)+(10-1)5]\)

\(=5[4+(9)\times(5)]\)

\(=5\times49=245\)

(ii) \(−37, −33, −29 ,…,\) to \(12\) terms

For this A.P.,

\(a = −37\)

\(d = a_2 − a_1 = (−33) −(-37)\)
(−37) = − 33 + 37 = 4

n = 12

We know that,

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

\(S_{12}=\frac{12}{2}[2(-37)+(12-1)4]\)

\(=6[-74+11\times4]\)

\(=6[-74+44]\)

\(=6(-30)=-180\)

(iii) \(6, 1.7, 2.8 ,…,\) to \(100\) terms

For this A.P.,

\(a = 0.6\)

\(d = a_2 − a_1 = 1.7 − 0.6 = 1.1\)
\(n = 100\)

We know that,

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

\(S_{100}=\frac{100}{2}[2(0.6)+(100-1)1.1]\)

\(=50[1.2+999\times(1.1)]\)

\(=50[1.2+108.9]\)

\(50[110.1]\)

\(=5505\)

(iv) \(\frac{1}{15},\frac{1}{12},\frac{1}{10}\)…. to \(11\) terms

For this A.P.,

\(a=\frac{1}{15}\)

\(n=11\)

\(d=a_2-a_1=\frac{1}{2}-\frac{1}{15}\)

\(=\frac{5-4}{60}=\frac{1}{60}\)

We know that,

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

\(S_n=\frac{11}{2}[2(\frac{1}{15}+(11-1)\frac{1}{60}]\)

\(=\frac{11}{2}[\frac{2}{15}+\frac{10}{15}]\)

\(=\frac{11}{2}[\frac{2}{15}+\frac{1}{6}]=\frac{11}{2}[\frac{4+5}{30}]\)

\(=(\frac{11}{2})(\frac{9}{30})=\frac{33}{20}\)

Question 2: Find the sums given below

(i) \(7+10\frac{1}{2}+14+………+84\)

(ii) \(34+32+30+…..+10\)

(iii) \(-5+(-8)+(-11)+………+(-230)\)

Answer: 

(i) \(7+10\frac{1}{2}+14+…….+84\)

For this A.P.,

\(a=7\)

\(l=84\)

\(d=a_2-a_1=10\frac{1}{2}-7\) \(=\frac{21}{2}-7=\frac{7}{2}\)

Let \(84\) be the n^th term of this A.P.

\(l=a+(n-1)d\)

\(84=7+(n-1)\frac{7}{2}\)

\(77=(n-1)\frac{7}{2}\)

\(22=n-1\)

\(n=23\)

We know that,

\(S_n=\frac{n}{2}(a+l)\)

\(S_n=\frac{23}{2}[7+84]\)

\(=\frac{23\times91}{2}=\frac{2098}{2}\)

\(=1046\frac{1}{2}\)

(ii) \(34+32+30+………..+10\)

For this A.P.,

\(a=34\)

\(d=a_2-a_1=32-34=-2\)

\(l=10\)

Let \(10\) be the n^th term of this A.P.

\(l=a+(n-1)d\)

\(10=34+(n-1)(-2)\)

\(-24=(n-1)(-2)\)

\(12=n-1\)

\(n=13\)

\(S_n=\frac{n}{2}(a+l)\)

\(=\frac{13}{2}(34+10)\)

\(=\frac{13\times44}{2}=13\times22\)

\(=286\)

(iii) \((-5)+(-8)+(-11)+…….+(-230)\)

For this A.P.

\(a=-5\)

\(l=230\)

\(d=a_2-a_1=(-8)-(-5)\)

\((-5)=-8+5=-3\)

Let \(-230\) be the n^th term of this AP.

\(l=a+(n-1)d\)

\(-230=-5+(n-1)(-3)\)

\(225=(n-1)(-3)\)

\(n-1)=75\)

\(n=76\)

And, \(S_n=\frac{n}{2}(a+l)\)

\(=\frac{76}{2}[(-5)+(-230)]\)

\(=38(-235)\)

\(=-8930\)

Question 3: In an A.P

(i) Given \(a = 5,\) \(d = 3\), \(an = 50\), find \(n\) and \(Sn\).
(ii) Given \(a = 7,\) \(a_{13} = 35\), find \(d\) and \(S_{13}\).
(iii) Given \(a_{12} = 37\), \(d = 3\), find \(a\) and \(S_{12}\).
(iv) Given \(a_3 = 15,\) \(S_{10} = 125\), find \(d\) and \(a_{10}\).

(v) Given \(d = 5\), \(S_9 = 75\), find \(a\) and \(a_9\).

(vi) Given \(a = 2\), \(d = 8\), \(Sn = 90\), find \(n\) and \(an\).

(vii) Given \(a = 8\), \(an = 62\), \(Sn = 210\), find \(n\) and \(d\).

(viii) Given \(an = 4\), \(d = 2\), \(Sn = − 14\), find \(n\) and \(a\).

(ix) Given \(a = 3\), \(n = 8\), \(S = 192\), find \(d\).

(x) Given \(l = 28\), \(S = 144\) and there are total \(9\) terms. Find \(a\).

Answer: 

Given that, \(a = 5\), \(d = 3\), \(an = 50\)

As \(an = a + (n − 1)d\)

\(∴ 50 = 5 + (n − 1)3\)
\(45 = (n − 1)3\)

\(15 = n − 1\)
\(n = 16\)

\(S_n=\frac{n}{2}[a+a_n]\)

\(S_{16}=\frac{16}{2}[+50]\)

\(=8\times55\)

\(=440\)

(ii) Given that, \(a=7\), \(a_{13}=35\)

As \(an=a+(n-1)d\)

\(∴ a_{13} = a + (13 − 1) d\)
\(35 = 7 + 12 d\)

\(35 − 7 = 12d\)
\(28 = 12d\)

\(d=\frac{7}{3}\)

\(S_n=\frac{n}{2}[a+a_n]\)

\(S_{13}=\frac{n}{2}[a+a_3]\)

\(=\frac{13}{2}[7+35]\)

\(=\frac{13\times42}{2}=13\times21\)

\(=272\)

(iii) Given that, \(a_{12}=37,\space d=3\)

As \(an=a+(n-1)d\)

\(a-{12}=a+(12-1)3\)

\(37=a+33\)

\(a=4\)

\(S_n=\frac{n}{2}[a+a_2]\)

\(S_n=\frac{12}{2}[4+37]\)

\(S_n=6(41)\)

\(S_n=246\)

(iv) Given that, \(a_3=15,\space S_{10}=125\)

As \(an=a+(n-1)d\)

\(a_3=a+(3-1)d\)

\(15=1+2d\)        …(i)

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

\(S_{10}=\frac{10}{2}[2a+(10-1)d]\)

\(125=5(2a+9d)\)

\(25=2a+9d\)

On multiplying equation (1) by 2, we obtain \(30=2a+4d\) (iii) On substracting equation (iii) from (ii), we obtain \(-5=5d\) \(d=-1\)

From equation (i),

\(15=a+2(-1)\)

\(15=a-2\)

\(a=17\)

\(a_{10}=a+(10-1)d\)

\(a_{10}=17+(9)(-1)\)

\(a_{10}=17-9=8\)

(v) Given that, \(d=5,\space S9=75\)

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

\(S_n=\frac{9}{2}[2a+(9-1)5]\)

\(75=\frac{9}{2}(2a+40)\)

\(25=3(a+20)\)

\(25=3a+60\)

\(3a=25-60\)

\(a=-\frac{35}{3}\)

\(an=a+(n-1)d\)

\(a_9=a+(9-1)(5)\)

\(=-\frac{35}{3}+895)\)

\(=-\frac{35}{3}=40\)

\(=\frac{-35+120}{3}=\frac{85}{3}\)

(vi) Given that, \(a=2,\space d=8,\space Sn=90\)

As \(S_n=\frac{n}{2}[2a+(n-1)d\)

\(90=\frac{n}{2}[4+(n-1)8]\)

\(90 = n [2 + (n − 1)4]\)
\(90 = n [2 + 4n − 4]\)

\(90 = n (4n − 2) = 4n^2 – 2n\)
\(4n^2 − 2n − 90 = 0\)
\(4n^2 − 20n + 18n − 90 = 0\)
\(4n (n − 5) + 18 (n − 5) = 0\)
\((n − 5) (4n + 18) = 0\)

Either \(n − 5 = 0\) or \(4n + 18 = 0\)

\(n=5\) or \(n=-\frac{18}{4}=-\frac{9}{2}\)

However, \(n\) can neither be \(n\) negative \(\) or fractional.

Therefore, \(n=5\)

\(an = a + (n − 1)d\)
\(a_5 = 2 + (5 − 1)8\)
\(= 2 + (4) (8)\)

\(= 2 + 32 = 34\)

(vii) Given that, \(1=8\), \(an=62\), \(Sn=210\)

\(S_n=\frac{n}{2}(a+a_n)\)

\(210=\frac{n}{2}[8+62]\)

\(210=\frac{n}{2}(70)\)

\(an=a+n(n-1)d\)

\(62=8+(6-1)d\)

\(62-8=5d\)

\(54=5d\)

\(d=\frac{54}{5}\)

(viii) Given that, \(an = 4\), \(d = 2\), \(Sn = −14\) \(an = a + (n − 1)d\)

\(4 = a + (n − 1)2\)
\(4 = a + 2n − 2\)
\(a + 2n = 6\)

\(a = 6 − 2n\)      … (i)

\(S_n=\frac{n}{2}[a+a_n]\)

\(-14=\frac{n}{2}[a+a_n]\)

\(-28=n(a+4)\)

\(-28=n(6-2n+4)\)

From equation (i) \(-28=n(-2n+10)\)

\(-28=-2n^2+10n\)

\(2n^2 − 10n − 28 = 0\)
\(n^2 − 5n −14 = 0\)

\(n^2 − 7n + 2n − 14 = 0\)

\(n (n − 7) + 2(n − 7) = 0\)
\((n − 7) (n + 2) = 0\)

Either \(n − 7 = 0\) or \(n + 2 = 0\)
\(n = 7\) or \(n = −2\)

However, \(n\) can neither be negative nor fractional.

Therefore, \(n = 7\)

From equation (i), we obtain

\(a = 6 − 2n\)
\(a = 6 − 2(7)\)
\(= 6 − 14\)

\(= −8\)

(ix) Given that, \(a=3\), \(n=8\), \(S=192\)

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

\(192=\frac{8}{2}[2\times3+(8-1)d]\)

\(192 = 4 [6 + 7d]\)

\(48 = 6 + 7d\)

\(42 = 7d\)

\(d = 6\)

(x) Given that, \(l = 28\), \(S = 144\) and there are total of \(9\) terms.

\(S_n=\frac{n}{2}(a+l)\)

\(144=\frac{9}{2}(a+28)\)

\((16)\times(2)=a+28\)

\(32=a+28\)

\(a=4\)

Question 4: How many terms of the AP. \(9, 17, 25 …\) must be taken to give a sum of \(636\)?

Answer:

Let there be n terms of this A.P.

For this A.P., \(a = 9\)

\(d = a_2 − a1 = 17 − 9 = 8\)

\(636 = n [9 + 4n − 4]\)
\(636 = n (4n + 5)\)
\(4n^2 + 5n − 636 = 0\)

\(4n^2 + 53n − 48n − 636 = 0\)

\(n (4n + 53) − 12 (4n + 53) = 0\)
\((4n + 53) (n − 12) = 0\)

Either \(4n + 53 = 0\) or \(n − 12 = 0\)

\(n=\frac{53}{4}\) or \(n=12\)

\(n\) cannot be \(-\frac{53}{4}\).

As the \(n\) number of term can neither be  negativa \(n\) or fractional, therefore, \(n=12\) only.

Question 5: The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer:

Given that,

\(a=5\)

\(S_n=\frac{n}{2}(a+l)\)

\(400=\frac{n}{2}(5+45)\)

\(400=\frac{n}{2}(50)\)

\(400=\frac{n}{2}(50)\)

\(n=16\)

\(l=45\)

\(45=5+(16-1)d\)

\(40=15d\)

\(d=\frac{40}{15}=\frac{8}{3}\)

Question 6: The first and the last term of an AP are 17 and 350 respectively. If the

common difference is 9, how many terms are there and what is their sum?

Answer:

Given that,

\(a = 17\)

\(d = 9\)

Let there be n terms in the A.P.

\(l = a + (n − 1) d\)

\(350 = 17 + (n − 1)9\)

\(333 = (n − 1)9\)
\((n − 1) = 37\)
\(n = 38\)

\(S_n=\frac{n}{2}(a+l)\)

\(⇒S_n=\frac{38}{2}(17+350)\)

\(⇒19(367)=6973\)

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

Question 7: Find the sum of first \(22\) terms of an AP in which \(d = 7\) and \(22^{nd}\) term is \(149\).

Answer:

\(d = 7\)

\(a_{22} = 149\)
\(S_{22} = ?\)

\(an = a + (n − 1)d\)

\(a_{22} = a + (22 − 1)d\)
\(149 = a + 21 × 7\)
\(149 = a + 147\)
\(a = 2\)

\(S_n=\frac{n}{2}(a+a_n)\)

\(=\frac{22}{2}(2+149)\)

\(=11(151)=1661\)

Question 8: Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Answer:

Given that,

\(a_2 = 14\)

\(a_3 = 18\)

\(d = a_3 − a_2 = 18 − 14 = 4\)
\(a_2 = a + d\)

\(14 = a + 4\)
\(a = 10\)

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

\(S_{51}=\frac{51}{2}[2\times10+(51-1)4]\)

\(=\frac{51}{2}[20+(50)(4)]\)

\(=\frac{51(220)}{2}=51(110)\)

\(=5610\)

Question 9: If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Answer:

Given that,

\(S_7=49\)

\(S_{17}=289\)

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

\(S_7=\frac{7}{2}[2a+(7-1)d]\)

\(49=\frac{7}{2}(2a+6d)\)

\(7=(1+3d)\)

\(a+3d=7\)    ..(i)

Similarly, \(S_{17}=\frac{17}{2}[2a+(17-1)d]\)

\(289=\frac{17}{2}[2a+16d]\)

\(17=(a+8d)\)

\(a+8d=17\)     …(ii) 

Subtracting equation (i) from equation (ii), \(5d = 10\)

\(d = 2\)

From equation (i),

\(a + 3(2) = 7\)

\(a + 6 = 7\)
\(a = 1\)

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

\(=\frac{n}{2}[2(1)+(n-1)(2)]\)

\(=\frac{n}{2}(2+2n-2)\)

\(=\frac{n}{2}(2n)\)

\(=n^2\)

Question 10: Show that \(a_1, a_2 … , an , …\) form an AP where an is defined as below:

(i) an = 3 + 4n

(ii) an = 9 − 5n

Answer:

(i) \(an = 3 + 4n\)

\(a_1 = 3 + 4(1) = 7\)

\(a_2 = 3 + 4(2) = 3 + 8 = 11\)
\(a_3 = 3 + 4(3) = 3 + 12 = 15\)
\(a_4 = 3 + 4(4) = 3 + 16 = 19\)

It can be observed that

\(a_2 − a_1 = 11 − 7 = 4\)
\(a_3 − a_2 = 15 − 11 = 4\)
\(a_4 − a_3 = 19 − 15 = 4\)

i.e.,  is same every time. Therefore, this is an AP with common difference as 4 and first term as \(7\).

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

\(S_{15}=\frac{15}{2}[2(7)+(15-1)4]\)

\(=\frac{15}{2}[(14)+56]\)

\(\frac{15}{2}(70)\)

\(=15\times35\)

\(=525\)

(ii) \(an = 9 − 5n\)

\(a_1 = 9 − 5 × 1 = 9 − 5 = 4\)

\(a_2 = 9 − 5 × 2 = 9 − 10 = −1\)
\(a_3 = 9 − 5 × 3 = 9 − 15 = −6\)
\(a_4 = 9 − 5 × 4 = 9 − 20 = −11\)

It can be observed that

\(a_2 − a_1 = − 1 − 4 = −5\)

\(a_3 − a_2 = − 6 − (−1) = −5\)
\(a_4 − a_3 = − 11 − (−6) = −5\)

i.e., \(ak + 1 − ak\) is same every time.

Therefore, this is an A.P. with common difference as \(−5\) and first term as \(4\).

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

\(S_{15}=\frac{15}{2}[2(4)+(15-1)(-5)]\)

\(=\frac{15}{2}(8-70)\)

\(=\frac{15}{2}(-62)=15(-31)\)

\(-465\)

Question 11: If the sum of the first n terms of an AP is \(4n − n^2\), what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3^rd, the10^th and the n^th terms.

Answer:

Given that,

\(Sn = 4n − n^2\)

First term, \(a = S_1 = 4(1) − (1)^2 = 4 − 1 = 3\)

Sum of first two terms \(= S_2\)

\(= 4(2) − (2)^2 = 8 − 4 = 4\)

Second term, \(a_2 = S_2 − S_1\) \(= 4 − 3 = 1 d\) \(= a_2 − a = 1 − 3\) \(= −2 an = a + (n − 1)d\)

\(= 3 + (n − 1) (−2)\)
\(= 3 − 2n + 2\)

\(= 5 − 2n\)

Therefore, \(a_3 = 5 − 2(3)\)

\(= 5 − 6 = −1 a_{10} = 5 − 2(10)\)

\(= 5 − 20 = −15\)

Hence, the sum of first two terms is 4.

The second term is 1. 3^rd, 10^th, and n^th terms are −1, −15, and 5 − 2n respectively.

 Question 12: Find the sum of first 40 positive integers divisible by 6.

Answer: The positive integers that are divisible by 6 are 6, 12, 18, 24 …

It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.

\(a = 6\)

\(d = 6\)
\(S_{40} =?\)

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

\(S_{40}=\frac{40}{2}[2(6)+(40-1)6]\)

\(= 20[12 + (39) (6)]\)
\(= 20(12 + 234)\)
\(= 20 × 246\)

\(= 4920\)

Question 13:

Find the sum of first 15 multiples of 8.

Answer:

The multiples of \(8\) are \(8, 16, 24, 32…\)

These are in an A.P., having first term as \(8\) and common difference as \(8\).

Therefore, \(a = 8\)

\(d = 8\)

\(S_15 =?\)

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

\(=\frac{15}{2}[2(8)+(15-1)8]\)

\(=\frac{15}{2}[16+14(8)]\)

\(=\frac{15}{2}(16+112)\)

\(=\frac{15(128)}{2}15\times64\)

\(=960\)

Question 14: Find the sum of the odd numbers between \(0\) and \(50\).

Answer:  The odd numbers between \(0\) and \(50\) are \(1, 3, 5, 7, 9 … 49\)

Therefore, it can be observed that these odd numbers are in an A.P.

\(A = 1\)

\(D = 2\)

\(l= 49\)

\(l= a + (n − 1) d\)

\(49 = 1 + (n − 1)2\)

\(48 = 2(n − 1)\)
\(n − 1 = 24\)
\(n = 25\)

\(S_n=\frac{n}{2}(a+l)\)

\(S_{25}=\frac{25}{2}(1+49)\)

\(=\frac{25(5)}{2}=(25)(25)\)

\(=625\)

Question 15: A contract on construction job specifies a penalty for delay of completion beyond a certain 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.

Answer: It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.

\(a = 200\)

\(d = 50\)

Penalty that has to be paid if he has delayed the work by \(30\) days \(=S_{30}\)

\(=\frac{30}{2}[2(200)+(30-1)50]\)

\(= 15 [400 + 1450]\)
\(= 15 (1850)\)

\(= 27750\)

Therefore, the contractor has to pay Rs \(27750\) as penalty.

Question 16: 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 of the prizes.

Answer:

Let the cost of 1^st prize be \(P\).

Cost of 2^nd prize \(= P − 20\)

And cost of 3^rd prize \(= P − 40\)

It can be observed that the cost of these prizes are in an A.P. having common difference as \(−20\) and first term as \(P\).

\(a = P\)

\(d = −20\)

Given that, \(S_7 = 700\)

\(\frac{7}{2}[2a+(7-1)d]=700\)

\(\frac{[2a+(6)(-20)]}{2}=100\)

\(a + 3(−20) = 100\)
\(a − 60 = 100\)

\(a = 160\)

Therefore, the value of each of the prizes was Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs. 60, and Rs. 40.

Question 17: In a school, students thought of planting 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 the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

Answer: It can be observed that the number of trees planted by the students is in an AP.

\(1, 2, 3, 4, 5………………..12\)
First term, \(a = 1\)

Common difference, \(d = 2 − 1 = 1 \)

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

\(S_{12}=\frac{12}{2}[2(1)+(12-1)(1)]\)

\(= 6 (2 + 11)\)

\(= 6 (13)\)

\(= 78\)

Therefore, number of trees planted by \(1\) section of the classes \(= 78\)

Number of trees planted by \(3\) sections of the classes \(= 3 × 78 = 234\)

Therefore, \(234\) trees will be planted by the students.

Question 18:

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Taken \(π=\frac{22}{7}\))

Answer: 

Semi-permeter of circle \(=πr\)

\(I_1=π(0.5)=\frac{π}{2}\) cm

\(I_2=π(1)=π\) cm

\(I_3=π(1.5)=\frac{3π}{2}\) cm

Therefore, \(I_1,I_2,I_3\), i.e. the lengths of the semi-circles are in A.P.,

\(\frac{π}{2},π,\frac{3π}{2},2π……\)

\(a=\frac{π}{2}\)

\(d=π-\frac{π}{2}=\frac{π}{2}\)

\(S_{13}=?\)

We know that the sum of \(n\) terms of an a A.P. is given by

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

\(=\frac{13}{2}[2(\frac{π}{2}+(13-1)(\frac{π}{2})]\)

\(=\frac{13}{2}[π+\frac{12π}{2}\)

\(=(\frac{13}{2})(7π)\)

\(=\frac{91π}{2}\)

\(=\frac{91\times22}{2\times7}=13\times11\)

\(=143\)

Therefore, the length of such spiral of thirteen consecutive semi-circles will be \(143\) cm.

Question 19: 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Answer: It can be observed that the numbers of logs in rows are in an A.P. \(20, 19, 18…\)

For this A.P.,

\(a = 20\)

\(d = a_2 − a_1 = 19 − 20 = −1\)

Let a total of \(200\) logs be placed in \(n\) rows.

\(Sn = 200\)

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

\(200=\frac{n}{2}[2(20)+(n-1)(-1)]\)

\(400 = n (40 − n + 1)\)
\(400 = n (41 − n)\)

\(400  = 41n − n^2\space n^2\)
\(− 41n + 400 = 0\)

\(n^2 − 16n − 25n + 400 = 0\)

\(n (n − 16) −25 (n − 16) = 0\)
\((n − 16) (n − 25) = 0\)

Either \((n − 16) = 0\) or \(n − 25 = 0\)
\(n = 16\) or \(n = 25\)

\(an = a + (n − 1)d\)

\(a_{16} = 20 + (16 − 1) (−1)\)
\(a_{16} = 20 − 15\)

\(a_{16} = 5\)

Similarly,

\(a_{25} = 20 + (25 − 1) (−1)\)
\(a_{25 }= 20 − 24\)

\(= −4\)

Clearly, the number of logs in 16^th row is 5.

However, the number of logs in 25^th row is negative, which is not possible.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^th row is 5.

Question 20: In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten 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 she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is \(2 × 5 + 2 ×(5 + 3)\)]

Answer:

The distances of potatoes are as follows. 5, 8, 11, 14…

It can be observed that these distances are in A.P. a = 5

\(d = 8 − 5 = 3\)

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

\(S_{10}=\frac{10}{2}[2(5)+(10-1)3]\)

\(= 5[10 + 9 × 3]\)

\(= 5(10 + 27) = 5(37)\)
\(= 185\)

As every time she has to run back to the bucket, therefore, the total
distance that the competitor has to run will be two times of it. Therefore, total distance that the competitor will run \(= 2 × 185 = 370\) m.

Alternatively,

The distances of potatoes from the bucket are \(5, 8, 11, 14…\)

Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, distances to be run are

\(10, 16, 22, 28, 34,……….\)
\(a = 10\)

\(d = 16 − 10 = 6\)
\(S_{10} =?\)

\(S_{10}=\frac{10}{2}[2\times10+(10-1)6]\)

Therefore, the competitor will run a total distance of \(370\) m.

Exercise 5.4

Question 1: Which term of the A.P. \(121, 117, 113 …\) is its first negative term?

[Hint: Find \(n\) for \(an < 0\)]

Answer:

Given A.P. is \(121, 117, 113 …\)

\(a = 121\)

\(d = 117 – 121 = -4\)

\(a_n = a + (n – 1) d\)

\(= 121 + (n – 1) (-4)\)

\(= 121 – 4_n + 4\)

\(= 125 – 4_n\)

We have to find the first negative term of this A.P.

Therefore, \(a<0\)

\(125-4_n<0\)

\(125<4_n\)

\(n>\frac{125}{4}\)

\(n>31.25\)

Therefore, 32^nd term will be the first negative term of this A.P.

Question 2: The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.

Answer:

We know that,

\(an = a + (n − 1) d\)
\(a_3 = a + (3 − 1) d\)
\(a_3 = a + 2d\)

Similarly, \(a_7 = a + 6d\)
Given that, \(a_3 + a_7 = 6\)

\((a + 2d) + (a + 6d) = 6\)
\(2a + 8d = 6\)

\(a + 4d = 3\)

\(a = 3 − 4d\)     …(i)

Also, it is given that \((a_3) × (a_7)\)

\(= 8 (a + 2d) × (a + 6d) = 8\)

From equation  (i)

(When \(d\) is \(\frac{1}{2}\)

\(a=3-4d\)

\(a=3-4(\frac{1}{2}\))

\(=3-2=1\)

(When \(d\) is \(-\frac{1}{2}\))

\(a=3-4(-\frac{1}{4})\)

\(a=3+2=5\)

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

(When \(a\) is \(1\) and \(d\) is \frac{1}{2}\))

\(S_{12}=\frac{16}{2}[2(1)+(16-1)(\frac{1}{2})]\)

\(=8[2+\frac{15}{2}]\)

\(=4(19)=76\)

(When \(a\) is \(5\) and \(d\) is \(-\frac{1}{2}\))

\(S_{16}=\frac{16}{2}[2(5)+(16-1)(-\frac{1}{2})]\)

\(=8[10+(15)(-\frac{1}{2})]\)

\(=8(\frac{5}{2})\)

\(=20\)

Question 3: A ladder has rungs 25 cm apart. (See figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are  \(\frac{2}{12}\)  m apart, what is the length of the wood required for the rungs?

[Hint: number of rungs \(=\frac{250}{25}\) ]

Answer: It is given that the rungs are 25 cm apart and the top and bottom rungs are \(2\frac{1}{2}\) m apart.

∴ Total number of rungs  \(\frac{2\frac{1}{2}\times100}{25}+1=\frac{250}{25}+1=11\)

Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.

The length of the wood required for the rungs equals the sum of all the terms of this A.P.

First term, \(a = 45\)

Last term, \(l = 25\)

\(n=11\)

\(S_n=\frac{n}{2}(a+l)\)

\(∴S_{10}=\frac{11}{2}(45+25)=\frac{11}{2}(70)=285\) cm

Therefore, the length of the wood required for the rungs is 385 cm.

Question 4: The houses of a row are number consecutively from 1 to 49. Show that there is a value of \(x\) such that the sum of numbers of the houses preceding the house numbered \(x\) is equal to the sum of the number of houses following it.

Find this value of \(x\).

[Hint \(Sx − 1 = S49 − Sx\)]

Answer:

The number of houses was \(1, 2, 3 … 49\)

It can be observed that the number of houses are in an A.P. having a as 1 and d also as \(1\).

Let us assume that the number of ^th house was like this.

Sum of \(n\) terms in an A.P. \(=\frac{n}{2}[2a+(n-1)d]\)

Sum of number of houses preceding \(x^{th}\) house \(=Sx-1\)

\(=\frac{(x-1)}{2} [2a+(x-1-1)d]\)

\(=\frac{x-1}{2} [2(1)+(x-2)(1)]\)

\(=\frac{x-1}{2}[2+x-2)\)

\(=\frac{(x)(x-1)}{2}\)

Sum of number of houses following \(x^{th}\) house \(= S_{49} – Sx\)

\(=\frac{49}{2} [2(1)+(49-1)(1)]-\frac{x}{2} [2(1)+(x-1)(1)]\)

\(=\frac{49}{2} (2+49-1)-\frac{x}{2}(2+x-1)\)

\(=(\frac{49}{2})(50)-\frac{x}{2}(2+x-1)\)

\(=25(49)-\frac{x(x+1)}{2}\)

It is given that these sums are equal to each other.

\(\frac{x(x-1)}{2}=25(49)-x(\frac{x+1}{2})\)

\(\frac{x^2}{2}-\frac{x}{2}=1225-\frac{x^2}{2}-\frac{x}{2}\)

\(x^2=1225\)

\(x=±35\)

However, the house numbers are positive integers.

The value of \(x\) will be 35 only.

Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

Question 5: A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of  \(\frac{1}{4}\) m and a tread of  \(\frac{1}{2}\) m (See figure) calculate the total volume of concrete required to build the terrace.

Answer: From the figure, it can be observe that

1^st step is  \(\frac{1}{2}\) m wide,

2^nd  step is \(1\) m wide,

3^rd  step is  \(\frac{3}{2}\)  m wide.

Therefore, the width of each step is increasing by  \(\frac{1}{3}\)  m each time whereas their height  \(\frac{1}{4}\)  m and length 50 m remains the same.

Therefore, the widths of these steps are   \(\frac{1}{2}, 1, \frac{3}{2}, 2……….\)

Volume of concrete in 1^st  step \(=\frac{1}{4}×\frac{1}{2}×5=\frac{25}{4}\)

Volume of concrete in 2^nd  step \(=\frac{1}{4}×1×50=\frac{25}{2}\)

Volume of concrete in 3^rd  step \(=\frac{1}{4}×\frac{3}{2}×50=\frac{75}{4}\)

It can be observed that the volumes of concrete in these steps are in an A.P.

\(\frac{25}{4},\frac{25}{2},\frac{75}{4}….\)

\(a=\frac{25}{4}\)

\(d=\frac{25}{2}-\frac{25}{4}=\frac{25}{4}\)

And \(S_n=\frac{n}{2}[2(\frac{25}{4})+(15-1)\frac{25}{4}\)

\(=\frac{15}{2}[\frac{25}{2}+\frac{(14)25}{4}]\)

\(\frac{15}{2}[\frac{25}{3}+\frac{175}{4}]\)

\(=\frac{15}{2}(100)=750\)

Volume of concrete required to build the terrace is 750 m³.