KT 10 Maths

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3

Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3

Question 1.
FInd the sum of the following APs:

Question 10.
Show that a1, a2, a3, … a11. … form an AP where a11 is defIned as below:
i) a = 3 + 4n
ii) a = 9 – 5n
Solution:
i) an= 3 + 4n
a1 = 3 + 4(1)
= 3 + 4
∴ an = 7
∴ a = 7
a1 = 3 + 4n
a2 = 3 + 4 × 2
= 3 + 8
∴ a2 = 11
an = 3 + 4n
a3 = 3 + 4 × 3
= 3 + 12
a3 = 15
∴ a1, a2, a3, ………….
7, 11, 15, ……….
d = a2 – a1 = 11 – 7 = 4
d = a3 – a2= 15 – 11 = 4
Here, the value of ‘d’ is constant.
∴ an = 3 + 4n forms an Arithmetic Progression.

ii) an = 9 – 5n
a1= 9 – 5 × 1
= 9 – 5
∴ a1 = 4
a1 = 9 – 5n
a1 = 9 – 5 × 2
= 9 – 10
∴ a2 = -1
an= 9 – 5n
a3 = 9 – 5 × 3
= 9 – 15
∴ a3 = -6
a1, a2, a3, …………… an
4, -1, -6, …………
d = a2 – a1 = -1 – 4 = -5
d=a3 – a2 = -6 – (-1) = -5
Here, the value of ‘d’ is constant.
∴ an = 9 – 5n form an Arithmetic Progression.

Question 11.
lf the sum of the first n terms of an AP is 4n – n2, what is the first term (that is SI)? What is the sum of first two termš? What is the second term? Similarly. find the 3rdrd the 10th and the nth terms.
Solution:
If Sn = 4n – n2, then
i) S1 = a =?
ii) S2 =?
iii) a2 =?
iv) a3 =?
v) a10 =?
vi) an =?
(i) Sn = 4n – n2
S1 = 4(1) – 12
= 4 – 1
S1 = 3
∴ S1 = a = 3.

(ii) S2 = 4n – n2
S2 = 4(2) – 22
= 8 – 4
S2 = 4
∴ S2 = 4

(iii) We have S2 = a1 + a2 = 4
= 3 + a2 = 4
∴ a2 = 4 – 3
∴a2 = 1

(iv) Sn = 4n – n2
S3= 4(3) – 32
= 12 – 9
S3 = 3
a1 + a2 + a3 = 3
3 + 1 + a3 = 3
4 + a3= 3
∴ a3 = 3 – 4
∴ a 3 = -1

(v) d = a3 – a3 = -1 -1 = -2
a10= a + 9d
= 3 + 9(-2)
= 3 – 18
a10= -15
∴ a10 = -15

(vi) an = a + (n – 1)d
= 3 + (n – 1)(-2)
= 3 + 2n + 2
∴ an = 5 – 2n
∴ an = 5 – 2n

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 cm. 1.0 cm. 1.5 cm, 2.0 cm as shows In fig. What Is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π=22/7)
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 8

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 (see the fig. given below). In how many rows are the 200 logs placed and how many logs are In the top row?
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 10

Question 20.
In a potato race, a bucket is placed at the starting point, which Is 5 m from the first potato. and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see fig, given below)
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 11
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops ¡tin 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 al) the potaotes 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)]
Solution:
Total distance competitor taken to pick up the first potato = 5 + 5m
= 2 × 5m.
= 10m.
Total distance taken by compeUtor to pick up the second potato
= 5 + 3 + 3 + 5m.
= 2 × 5 + 2 × 3
= 2(5 + 3)
= 2 × 8
= 16m.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3, drop a comment below and we will get back to you at the earliest.

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