Ncert Solutions For Class 10 Maths Ex 5.1

Ncert Solutions For Class 10 Maths Chapter 5 Ex 5.1

General Form of an A.P:

Let us consider a series with n number of terms:

a1 + a2 + a3 + a4 + a5 + a6 + a7 ……………………………..  = an

The above series of numbers is said to form an Arithmetic Progression (A.P) if,

a2 – a1 = a3 – a2 = a4 – a3 = …………………………. an – an-1 = d

Here, d = common difference of the A.P and it can be positive negative or zero.

Thus the general form of an A.P can be written as:

a, a+ d, a+ 2d, a+ 3d, a+ 4d, a+ 5d …………………………. a + (n – 1)d

Where a = First Term

And,    d = Common Difference

 

Q-1: From the given A.P write the first term (a) and common difference (d). 

(i) 2, 4, 6, 8, 10…..

Sol.

Here, a = 2

Since d = a2 – a1

Therefore, d = 4 – 2 = 2

Q-2: Check whether the following list of numbers form an A.P? If they form an A.P then write next three terms.

(i) 5, 25, 45, 65………

Sol.

Here, a = 5

Since for an A.P:     a2 – a1 = a3 – a2 = a4 – a3

Therefore,      25 – 5 = 45 – 25 = 65 – 45 = 20

Hence the given number forms an A.P with d = 20

Therefore next three terms of this A.P are: 65 + 20 = 85

85 + 20 =105

105 + 20 = 125

 

(ii) 12,14,18

Sol.

Here, a = 12

Since for an A.P:      a2 – a1 = a3 – a2 = a4 – a3

Therefore, 14121814

Hence the given numbers does not form an A.P

The nth term of an A.P is given by:

           an = a + (n – 1)×d

Here,  a = first term of an A.P and d = common difference

Example-1: Let us consider an A.P 2, 7, 12, 17……. Find its 9th term.

Sol.

Here,                   a = 2

d = 7 – 2 = 17 – 12 = 5

n = 9

Therefore,   a9 = 2+ (9 – 1)5

a9 = 42

Therefore, 9th term of this A.P = 4

Example-2 Fourth term of an A.P is 12 and Sixth term of an A.P is 18. Determine the A.P

Sol.

Given, a4 = 12 and a6 = 18

Let a be the first term and d be the common difference of an A.P

Then               12 = a + (4 – 1)d

Therefore,     12 = a + 3d . .  . . (1)

And                 18 = a + (6 – 1)d

Therefore      18 = a + 5d . . . . . . . (2)

Now, equation (2) – equation (1)

18 – 12 = 5d – 3d    Therefore d = 3

Putting d = 3 in equation (2) we get

18 = a + 5×(3)

Therefore a = 3

Hence, the required A.P is 3, 6, 9, 12, 15, 18, 21………

 

Example -3 Determine which term of the following A.P : 14, 9, 4 . . . . . . is -96 ?

Sol.

From the given A.P,

a = 14,

d = -5,

an = -96,

n =??

Since,            an = a + (n – 1)d

Therefore   -96 = 14 + (n – 1)×-5

(or)                5n = -96 -14 -5

Therefore       n = -23 (neglecting –ve sign)

Therefore, 23rd term of this A.P is -96.

 

Example-4 Check whether 328 is the term in the following A.P: 76, 97, 118, 139………

Sol.

From the given A.P:

a = 76,

d = 97 – 76 = 21,

an = 328

Since,                                       an = a + (n – 1)d

Therefore                            328 = 76 + (n – 1)×21

(or)                                       21n = 328 + 21 – 76

(or)                                           n = 13

Since n is a +ve integer

Therefore 328 is the 13th term of this given A.P

 

Example -5 Find how many two digit numbers are divisible by 5.?

Sol.

Two digit numbers that are divisible by 5 are: 10, 15, 20, 25…………………95

Now, from the given A.P

a = 10

d = 5

an = 95

n = ??

Since,         an = a + (n – 1)d

Therefore 95 = 10 + (n – 1)×5

(or)         5n = 95 – 10 + 5

n = 18

Therefore there are 18 two digit numbers that are divisible by 5.

Example-6 From the given A.P: 15, 8, 1, -6………….. -111. Find 8th term from last term.

Sol.

On reversing the given A.P:

a = -111 (in this case, last term will now become its first term)

d = 7 (If we go from first term to last term then d = -7 (8 – 15), so if we reverse this A.P d will become +7)

n = 8

a8 = ??

Since,          an = a + (n – 1)d

Therefore, a8  = -111 + (8 – 1)×7 = -111 + 49

a8 = -62

Therefore, 8th term from last term = -62

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