NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Exercise 5.1

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

Sol.1) i) It can be observed that
Taxi fare for 1st 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 litres.
In each stroke, the vacuum pump removes 1/4
of air remaining in the cylinder at a time.
In other words, after every stroke, only 1 − (1/4) = 3/4𝑡ℎ part of air will remain.
Therefore, volumes will be 𝑉, 3𝑉/4, (3𝑉/4)2, (3𝑉/4)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 years, our money will be 𝑃 (1 + 𝑟/1000)𝑛
therefore, after every year, our money will be
10000 (1 + 8/100) , 10000 (1 + 8/100)2, 10000 (1 + 8/100)3….
Clearly, adjacent terms of this series do not have the same difference between them.
Therefore, this is not an A.P.

Q.2) Write first four terms of the A.P. when the first term a and the common differenced are given as follows
(i) a = 10, d = 10 (ii) a = -2, d = 0 (iii) a = 4, d = - 3
(iv) a = -1 d = 1/2 (v) a = - 1.25, d = - 0.25
Sol.2) (i) 𝑎 = 10, 𝑑 = 10
Let the series be 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 , 𝑎5 …
𝑎1 = 𝑎 = 10
𝑎2 = 𝑎1 + 𝑑 = 10 + 10 = 20
𝑎3 = 𝑎2 + 𝑑 = 20 + 10 = 30
𝑎4 = 𝑎3 + 𝑑 = 30 + 10 = 40
𝑎5 = 𝑎4 + 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, 𝑑 = 0
Let the series be 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 …
𝑎1 = 𝑎 = −2
𝑎2 = 𝑎1 + 𝑑 = − 2 + 0 = − 2
𝑎3 = 𝑎2 + 𝑑 = − 2 + 0 = − 2
𝑎4 = 𝑎3 + 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, 𝑑 = − 3
Let the series be 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 …
𝑎1 = 𝑎 = 4
𝑎2 = 𝑎1 + 𝑑 = 4 − 3 = 1
𝑎3 = 𝑎2 + 𝑑 = 1 − 3 = − 2
𝑎4 = 𝑎3 + 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, 𝑑 = 1/2
Let the series be 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 …
𝑎1 = 𝑎 = −1

First four terms of this A.P. will be -1, -1/2, 0 and 1/2.
(v) 𝑎 = − 1.25, 𝑑 = − 0.25
Let the series be 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4…
𝑎1 = 𝑎 = − 1.25
𝑎2 = 𝑎1 + 𝑑 = − 1.25 − 0.25 = − 1.50
𝑎3 = 𝑎2 + 𝑑 = − 1.50 − 0.25 = − 1.75
𝑎4 = 𝑎3 + 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.

Q.3) For the following A.P.s, write the first term and the common difference.
(i) 3, 1, - 1, - 3 … (ii) - 5, - 1, 3, 7 … (iii) 1/3, 5/3, 9/3, 13/3 ....
(iv) 0.6, 1.7, 2.8, 3.9 …
Sol.3) (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) 1/3, 5/3, 9/3, 13/3 ....
Here, first term, a = 1/3
Common difference, d = second term – first term = 5/3 – 1/3 = 4/3
(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

Q.1) Fill in the blanks in the following table, given that a is the first term, d the common difference and an the 𝑛𝑡ℎ term of the A.P.

Sol.1) 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

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
𝑑 = 18/9 = 2
Hence, common difference, 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

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) = 22.5/2.5
𝑛 − 1 = 9
𝑛 = 10
Hence, 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 × 0
𝑎𝑛 = 3.5
Hence, 𝑎𝑛 = 3.5

Q.2) Choose the correct choice in the following and justify
(i). 30th term of the A.P: 10, 7, 4, ..., is
(A). 97 (B). 77 (C). − 77 (D). – 87
(ii). 11th term of the A.P. is
(A). 28 (B). 22 (C). − 38 (D).−48(1/2)
Sol.2) (i) Given that
A.P. 10, 7, 4, ...
First term, a = 10
Common difference, 𝑑 = 𝑎2 – 𝑎1 = 7 − 10 = −3
We know that, 𝑎𝑛 = 𝑎 + (𝑛 − 1) 𝑑
𝑎30 = 10 + (30 − 1) (−3)
𝑎30 = 10 + (29) (−3)
𝑎30 = 10 − 87 = −77
Hence, the correct answer is C.
(ii) Given that, A.P. −3, − 1/2 , 2, … …
First term 𝑎 = −3
Common difference, 𝑑 = 𝑎2 – 𝑎1
= − 1/2 − (−3)
= − 1/2 + 3 = 5/2
We know that,
𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑
𝑎11 = −3 + (11 − 1) (5/2)
𝑎11 = −3 + (10) (5/2)
𝑎11 = −3 + 25
𝑎11 = 22
Hence, the answer is B.

Q.3) In the following APs find the missing term in the boxes
i) 2, __, 26 ii) ___, 13, __,3 iii) 5,___, ___,9,1/2
iv) -4,___, ___,___, ___,6 v) __,38,___,___,___, -22
Sol.3) For this A.P., a = 2 and 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) For this A.P., 𝑎2 = 13 and 𝑎4 = 3
We know that, an = a + (n − 1) d
𝑎2 = 𝑎 + (2 − 1) 𝑑
13 = 𝑎 + 𝑑 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (𝐼)
𝑎4 = 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
𝑎3 = 18 + (3 − 1) (−5)
= 18 + 2 (−5) = 18 − 10 = 8
Therefore, the missing terms are 18 and 8 respectively.

iii) For this A.P.,
𝑎 = 5 and 𝑎4 = 19/2
We know that, 𝑎𝑛 = 𝑎 + (𝑛 – 1)𝑑
𝑎4 = 𝑎 + (4 − 1)𝑑 
19/2 = 5 + 3𝑑
19/2 − 5 = 3𝑑
3𝑑 = 9/2
𝑑 = 3/2
𝑎2 = 𝑎 + (2 − 1)𝑑
𝑎2 = 5 + 3/2
𝑎2 = 13/2
𝑎3 = 𝑎 + (3 − 1)𝑑
𝑎3 = 5 + 2 × 3/2
𝑎3 = 8
Therefore, the missing terms are
13/2 and 8 respectively.

(iv) Here, a = - 4 and 𝑎6 = 6 
common difference can be calculated as follows:
𝑎6 = 𝑎 + 5𝑑
6 = −4 + 5𝑑
5𝑑 = 6 + 4 = 10
𝑑 = 2
The second, third, fourth and fifth terms of this AP can be calculated as follows:
𝑎2 = 𝑎 + 𝑑 = − 4 + 2 = − 2
𝑎3 = 𝑎 + 2𝑑 = − 4 + 4 = 0
𝑎4 = 𝑎 + 3𝑑 = − 4 + 6 = 2
𝑎5 = 𝑎 + 4𝑑 = − 4 + 8 = 4
Thus, the given AP can be written as: - 4, - 2, 0, 2, 4, 6

(v) (...), 38, (...), (...), (...), - 22
Let us take 38 as the first term and – 22 as the 5th term
Using this, common difference can be calculated as follows:
𝑎5 = 𝑎 + 4𝑑
− 22 = 38 + 4𝑑
4𝑑 = − 22 – 38 = − 60
𝑑 = − 15
If 38 is the second term, then first term can be calculated as follows:
𝑎 = 𝑎2 – 𝑑 = 38 + 15 = 53
Third, fourth and fifth terms can be calculated as follows:
𝑎3 = 𝑎 + 2𝑑 = 53 + 2(− 15) = 53 – 30 = 23
𝑎4 = 𝑎 + 3𝑑 = 53 – 45 = 8
𝑎5 = 𝑎 + 4𝑑 = 53 – 60 = − 7
So, the AP can be written as: 53, 38, 23, 8, - 7, - 22

Q.4) Which term of the AP: 3, 8, 13, 18, ……………… is 78?
Sol.4) Given, 𝑎 = 3, 𝑑 = 𝑎2 – 𝑎1 = 8 – 3 = 5, 𝑎𝑛 = 78, 𝑛 = ?
We know that an = 𝑎 + (𝑛 – 1)𝑑
78 = 3 + (𝑛 – 1)5
(𝑛 – 1)5 = 78 – 3 = 75
𝑛 – 1 = 15
𝑛 = 15 + 1 = 16
Thus, 78 is the 16th term of given AP.

Q.5) Find the number of terms in each of the following Aps
Sol.5) (i) 7, 13, 18, ……, 205
Here, 𝑎 = 7, 𝑑 = 6, 𝑎𝑛 = 205, 𝑛 = ?
We know that 𝑎𝑛 = 𝑎 + (𝑛 – 1)𝑑
205 = 7 + (𝑛 – 1)6
(𝑛 – 1)6 = 205 – 7 = 198
𝑛 – 1 = 33
𝑛 = 34
Thus, 205 is the 34th of term of this AP.
(ii) 18, 15.5, 13, …………, - 47
Here, 𝑎 = 18, 𝑑 = 15.5 – 18 = − 2.5
We know that 𝑎𝑛 = 𝑎 + (𝑛 – 1)𝑑
− 47 = 18 + (𝑛 – 1)(− 2.5)
(𝑛 – 1)(− 2.5) = − 47 – 18 = − 65
𝑛 – 1 = 26
𝑛 = 27
Thus, - 47 is the 27th term of this AP.

Q.6) Check whether – 150 is a term of the AP; 11, 8, 5, 2, ……………
Sol.6) Here, 𝑎 = 11, 𝑑 = 8 – 11 = − 3, 𝑎𝑛 = − 150, 𝑛 = ?
We know that 𝑎𝑛 = 𝑎 + (𝑛 – 1)𝑑
− 150 = 11 + (𝑛 – 1)(− 3)
(𝑛 – 1)(−3) = − 150 – 11 = − 161
𝑛 – 1 = 161/3
It is clear that 161 is not divisible by three and we shall get a fraction as a result. But
number of term cannot be a fraction.
Hence, - 150 is not a term of the given AP.

Q.7) Find the 31st term of an AP whose 11th term is 38 and 16th term is 73
Sol.7) Given, 𝑎11 = 38 and 𝑎16 = 73
We know that 𝑎𝑛 = 𝑎 + (𝑛 – 1)𝑑
Hence, 𝑎11 = 𝑎 + 10𝑑 = 38
And, 𝑎16 = 𝑎 + 15𝑑 = 73
Subtracting 11th term from 16th term, we get following:
𝑎 + 15𝑑 – 𝑎 – 10𝑑 = 73 – 38
5𝑑 = 35
𝑑 = 7
Substituting the value of d in 11th term we get;
𝑎 + 10 × 7 = 38
𝑎 + 70 = 38
𝑎 = 38 – 70 = − 32
Now 31st term can be calculated as follows:
𝑎31 = 𝑎 + 30𝑑
= − 32 + 30 × 7
= − 32 + 210 = 178

Q.8) An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Sol.8) Given, 𝑎3 = 12 and 𝑎50 = 106
𝑎3 = 𝑎 + 2𝑑 = 12
𝑎50 = 𝑎 + 49𝑑 = 106
Subtracting 3rd term from 50th term, we get;
𝑎 + 49𝑑 – 𝑎 – 2𝑑 = 106 – 12
47𝑑 = 94
𝑑 = 2
Substituting the value of d in 12th term, we get;
𝑎 + 2 × 2 = 12
𝑎 + 4 = 12
𝑎 = 8
Now, 29th term can be calculated as follows:
𝑎29 = 𝑎 + 28𝑑
= 8 + 28 × 2
= 8 + 56 = 64

Q.9) If the 3rd and the 9th term of an AP are 4 and – 8 respectively. Which term of this AP is zero?
Sol.9) Given, 𝑎3 = 4 𝑎𝑛𝑑 𝑎9 = − 8
𝑎3 = 𝑎 + 2𝑑 = 4
𝑎9 = 𝑎 + 8𝑑 = − 8
Subtracting 3rd term from 9th term, we get;
𝑎 + 8𝑑 – 𝑎 – 2𝑑 = − 8 – 4 = − 12
6𝑑 = − 12
𝑑 = − 2
Substituting the value of d in 3rd term, we get;
𝑎 + 2(−2) = 4
𝑎 – 4 = 4
𝑎 = 8
Now; 0 = 𝑎 + (𝑛 – 1)𝑑
0 = 8 + (𝑛 – 1)(− 2)
(𝑛 – 1)(− 2) = − 8
𝑛 – 1 = 4
𝑛 = 5
Thus, 5th term of this AP is zero.

Q.10) The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Sol.10) Tenth and seventeenth terms of this AP can be given as follows:
𝑎10 = 𝑎 + 9𝑑
𝑎17 = 𝑎 + 16𝑑
Subtracting 10th term from 17th term, we get;
𝑎 + 16𝑑 – 𝑎 – 9𝑑 = 7
7𝑑 = 7
𝑑 = 1

Q.11) Which term of the 𝐴𝑃: 3, 15, 27, 39, …. will be 132 more than its 54th term.
Sol.11) Here, 𝑎 = 3, 𝑑 = 15 – 3 = 12
54th term can be given as follows:
𝑎54 = 𝑎 + 53𝑑
= 3 + 53 × 12
= 3 + 636 = 639
So, the required term = 639 + 132 = 771
771 = 𝑎 + (𝑛 – 1)𝑑
771 = 3 + (𝑛 − 1)12
(𝑛 – 1)12 = 771 – 3 = 768
𝑛 – 1 = 64
𝑛 = 65
Thus, the required term is 65th term

Q.12) Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Sol.12) Since, both the APs have same common difference, thus in this case the difference between each corresponding terms will be 100.

Q.13) How many three digit numbers are divisible by 7?
Sol.13) Since, 100 is the smallest three digit number and it gives a reminder of 2 when divided by 7, therefore, 105 is the smallest three digit number which is divisible by 7.
Since, 999 is greatest three digit number, and it gives a reminder of 5,
thus 999 – 5 = 994 will be the greatest three digit number which is divisible by 7.
Therefore, here we have,
First term (𝑎) = 105,
The last term (𝑎𝑛) = 994
The common difference = 7
We know that, 𝑎𝑛 = 𝑎 + (𝑛 – 1)𝑑
994 = 105 + (𝑛 – 1)7
(𝑛 – 1)7 = 994 – 105 = 889
𝑛 – 1 = 127
𝑛 = 128
Thus, there are 128 three digit numbers which are divisible by 7.

Q.14) How many multiples of 4 lie between 10 and 250?
Sol.14) 2 is the first number after 10 which is divisible by 4.
Since, 250 gives a remainder of 2 when divided by 4, thus 250 – 2 = 248 is the greatest number less than 250 which is divisible by 4.
Here, we have first term (𝑎) = 12, last term (𝑛) = 248
and common difference (𝑑) = 4
Thus, number of terms (𝑛) =?
We know that, 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑
248 = 12 + (𝑛 – 1)4
(𝑛 – 1)4 = 248 – 12 = 236
𝑛 – 1 = 59
𝑛 = 60
Thus, there are 60 numbers between 10 and 250 that are divisible by 4.

Q.15) For what value of n, are the nth terms of two APs; 63, 65, 67, ………. and 3, 10, 17, ……. equal.
Sol.15) In first 𝐴𝑃: 𝑎 = 63, 𝑑 = 2
In second 𝐴𝑃: 𝑎 = 3, 𝑑 = 7
As per question:
63 + (𝑛 – 1) 2 = 3 + (𝑛 – 1) 7
⇒ 63 – 3 + (𝑛 – 1) 2 = (𝑛 – 1) 7
⇒ 60 + 2𝑛 – 2 = 7𝑛 – 7
⇒ 2𝑛 + 58 = 7𝑛 – 7
⇒ 2𝑛 + 58 + 7 = 7𝑛
⇒ 2𝑛 + 65 = 7𝑛
⇒ 7𝑛 – 2𝑛 = 65
⇒ 5𝑛 = 65
⇒ 𝑛 = 65/5 = 13
Thus, for the 13 value of 𝑛, 𝑛𝑡ℎ term of given two APs will be equal

Q.16) Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Sol.16) Given 𝑎3 = 16 and 𝑎7 – 𝑎5 = 12
𝑎3 = 𝑎 + 2𝑑 = 16
𝑎5 = 𝑎 + 4𝑑
𝑎7 = 𝑎 + 6𝑑
As per question;
𝑎 + 6𝑑 – 𝑎 – 4𝑑 = 12
2𝑑 = 12
𝑑 = 6
Substituting the value of d in third term, we get;
𝑎 + 2 × 12 = 16
𝑎 + 24 = 16
𝑎 = 16 – 24 = − 8
Thus, the AP can be given as follows:
-8, 4, 16, 28, 40, ………..

Q.17) Find the 20th term from the last term of the AP; 3, 8, 13, …………….253.
Sol.17) 𝑎 = 3, 𝑑 = 5
Now, 253 = 𝑎 + (𝑛 + 1) 𝑑
⇒ 253 = 3 + (𝑛 − 1) × 5
⇒ 253 = 3 + 5𝑛 – 5 = – 2
⇒ 5𝑛 = 253 + 2 = 255
⇒ 𝑛 = 255/5 = 51
Therefore, 20th term from the last term = 51 – 19 = 32
𝑎32 = 𝑎 + 31𝑑
= 3 + 31 × 5
= 3 + 155 = 158
Thus, required term is 158.

Q.18) The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Sol.18) Given, 𝑎8 + 𝑎4 = 24 and 𝑎10 + 𝑎6 = 44
𝑎8 = 𝑎 + 7𝑑
𝑎4 = 𝑎 + 3𝑑
As per question;
𝑎 + 7𝑑 + 𝑎 + 3𝑑 = 24
2𝑎 + 10𝑑 = 24
𝑎 + 5𝑑 = 12 …………(1)
𝑎10 = 𝑎 + 9𝑑
𝑎 + 9𝑑 + 𝑎 + 5𝑑 = 44
2𝑎 + 14𝑑 = 44
𝑎 + 7𝑑 = 22 ………….(2)
Subtracting equation (1) from equation (2);
𝑎 + 7𝑑 – 𝑎 – 5𝑑 = 22 – 12
2𝑑 = 10
𝑑 = 5
Substituting the value of d in equation (1), we get;
𝑎 + 5 × 5 = 12
𝑎 + 25 = 12
𝑎 = − 13
Hence, first three terms of AP: – 13, – 8, – 3

Q.19) Subha 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?
Sol.19) Here, 𝑎 = 5000, 𝑑 = 200 and 𝑎𝑛 = 7000
We know, 𝑎𝑛 = 𝑎 + (𝑛 – 1)𝑑
7000 = 5000 + (𝑛 – 1)200
(𝑛 − 1)200 = 7000 – 5000
(𝑛 – 1)200 = 2000
𝑛 – 1 = 10
𝑛 = 11
Thus, 1995 + 10 = 2005
Hence, his salary reached at Rs. 7000 in 2005.

Q.20) Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs. 20.75, find n.
Sol.20) Here, a = 5, d = 1.75 and an = 20.75
We know, 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑
20.75 = 5 + (𝑛 – 1)1.75
(𝑛 – 1)1.75 = 20.75 – 5
(𝑛 – 1)1.75 = 15.75
𝑛 – 1 = 15.75/1.75 = 9
𝑛 = 10

Exercise 5.3

Q.1) Find the sum of the following APs:

Sol.1) (i) 2, 7, 1, 2, ………………….. to 10 terms
Here, 𝑎 = 2, 𝑑 = 5 and 𝑛 = 10
Sum of 𝑛 terms can be given as follows:
𝑆 = 𝑛/2 [2𝑎 + (𝑛 − 1)𝑑]
𝑆10 = 10/2 [2 × 2 + (10 − 1)5]
= 5(4 + 45)
= 5 × 49 = 245
Thus, sum of the 10 terms of given 𝐴𝑃 (𝑆𝑛) = 245

(ii) – 37, – 33, – 29, ………………..to 12 terms
Here, 𝑎 = − 37, 𝑑 = 4 and 𝑛 = 12
Sum of 𝑛 terms can be given as follows:
𝑆 = 𝑛/2
[2𝑎 + (𝑛 − 1)𝑑]
𝑆12 = 12/2
[2(−37) + 11 × 4]
= 6(−74 + 44)
= 6(−30) = −180
Thus, sum of the 12 terms of given 𝐴𝑃 (𝑆𝑛) = – 180

(iii) 0.6, 1.7, 2.8, ……………… to 100 terms
Here, 𝑎 = 0.6, 𝑑 = 1.1 and 𝑛 = 100
Sum of 𝑛 terms can be given as follows:
𝑆 = 𝑛/2 [2𝑎 + (𝑛 − 1)𝑑]
𝑆100 = 100/2 [2 × 0.6 + 99 × 1.1]
= 50(1.2 + 108.9)
= 50 × 110.1 = 5505
Thus, sum of the 100 term of given 𝐴𝑃 (𝑆𝑛) = 5505

Q.2) Find the sums given below:
Sol.2) (i) 7 + 10.5 + 14 + ….. + 84
Here, 𝑎 = 7, 𝑑 = 3.5 and last term = 84
Number of terms can be calculated as follows;
𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑
84 = 7 + (𝑛 − 1)3.5
(𝑛 − 1)3.5 = 84 − 7
𝑛 − 1 = 77/3.5 = 22
𝑛 = 23
Sum of 𝑛 terms can be given as follows:
𝑆 = 𝑛/2 [2𝑎 + (𝑛 − 1)𝑑]
= 23/2 [2 × 7 + 22 × 33.5]
= 23/2 (14 + 77)
= 23/2 × 91 = 2093/2
= 1046(1/2)

(ii) 34 + 32 + 30 + …. + 10
Here, 𝑎 = 34, 𝑑 = − 2 and last term = 10
Number of terms can be calculated as follows:
𝑎𝑛 = 𝑎 + (𝑛 – 1)𝑑
10 = 34 + (𝑛 – 1)(− 2)
10 = 34 – (𝑛 – 1)(2)
(𝑛 – 1)2 = 34 – 10 = 24
𝑛 – 1 = 12
𝑛 = 13
Sum of 𝑛 terms can be given as follows:
𝑆 = 𝑛/2 [2𝑎 + (𝑛 − 1)𝑑]
= 13/2 [2 × 34 + 12(−2)]
= 13/2 (68 − 24)
= 13/2 × 44 = 286
Thus sum of given 𝐴𝑃 = 286

(iii) – 5 + (-8) + (- 11) + …… + (- 230)
Here, 𝑎 = − 5, 𝑑 = − 3 and last term = − 230
Number of terms can be calculated as follows:
𝑎𝑛 = 𝑎 + (𝑛 – 1)𝑑
− 230 = − 5 + (𝑛 – 1)( − 3)
− 230 = − 5 – (𝑛 – 1)3
(𝑛 – 1)3 = − 5 + 230 = 225
𝑛 – 1 = 75
𝑛 = 76
Sum of 𝑛 terms can be given as follows:
𝑆 = 𝑛/2 [2𝑎 + (𝑛 − 1)𝑑]
= 76/2 [2(−5) + 75(−3)]
= 38(−10 − 225)
= 38(−235) = −8930
Thus sum of given 𝐴𝑃 = −8930

Q.3) In an AP:
(a) Given 𝑎 = 5, 𝑑 = 3, 𝑎𝑛 = 50, find 𝑛 and 𝑆𝑛.
Number of terms can be calculated as follows:
𝑎𝑛 = 𝑎 + (𝑛 – 1)𝑑
50 = 5 + (𝑛 – 1)3
(𝑛 – 1)3 = 50 – 5 = 45
𝑛 – 1 = 15
𝑛 = 16
um of 𝑛 terms can be given as follows:
𝑆 = 𝑛/2 [2𝑎 + (𝑛 − 1)𝑑]
= 16/2 [2 × 5 + 15 × 3]
= 8(10 + 45)
= 8 × 55 = 440
Thus 𝑛 = 16 and sum = 440

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