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ncert solutions for class 7 maths is a must for CBSE Class 7 students. Mathematics is one of those subjects that requires constant attention. At Entrancei we have math experts on our team.

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Published by abhi.012101, 2021-03-25 06:01:01

ncert solutions for class 7 maths

ncert solutions for class 7 maths is a must for CBSE Class 7 students. Mathematics is one of those subjects that requires constant attention. At Entrancei we have math experts on our team.

Keywords: ncert solutions for class 7 maths,ncert solutions for class 9 maths,ncert solutions for class 8 maths

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ARITHMETIC PROGRESSION

1. Arithmetic Progression (A.P.).
Arithmetic Progression (or Arithmetic sequence) is a sequence in which the difference
of a term and its predecessor is always constant, i.e., an − an−1 = d, where d is the
common difference. The successive terms of an A.P. are
a, a + d, a + 2d, a + 3d, a + 4d ..........

• Finite A.P.: An A.P. containing finite number of terms is called finite A.P.

e.g. 147, 149, 151 ………………….. 163.

• Infinite A.P.: An A.P. containing infinite terms is called infinite A.P.

e.g. 6, 9, 12, 15 ………………………..

For Example:

Which of the following are A.P’s? If they form an A.P., find the common difference d
and the next three terms after last given term.

(i) 3, 6, 9, ... (ii) 12, 52, 72, 73, …

(i) 3, 6, 9, ...

a2 − a1 = 6 − 3 = 3 ( 2 − 1)
a3 − a2 = 9 − 6 = 3 ( 3 − 2 )
..............................................................
..............................................................
As the common difference between any two consecutive terms is not the same

 The given sequence is not an A.P.

(ii) 12,52,72,73,....

a2 − a1 = 52 − 1 = 25 − 1 = 24
a3 − a2 = 72 − 52 = 49 − 25 = 24

a4 − a3 = 73 − 49 = 24
.................................................
.................................................

As the common difference between any two consecutive terms is the same

 The given sequence is an A.P.

Next three terms are:

a5 = 73 + 24 = 97 a6 = 97 + 24 = 121

a7 = 121+ 24 = 145

2. General Term.
General term or nth term of an A.P. is an = a + (n − 1)d, where a = first term, d =

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common difference.

• The formula an = a + (n − 1)d contains four quantities an , a, n and d. Three
quantities being given, the fourth can be found by using above relation.

For Example:

Find the 18th term and nth term for the sequence 7, 4, 1, −2, −5.
Here a = 7
and d = a2 − a1

=4−7=−3
n = 18

an = a + (n − 1)d
a18 = 7 + (18 − 1) −3 = 7 + 17  −3

= 7 − 51 = −44

an = a + (n − 1)d

= 7 + (n − 1) (−3)
= 7 − 3n + 3
= 10 − 3n

• If only two quantities are given, two conditions (equations) in the problem should
be given. Therefore, to determine these two unknowns, we have to solve both the
conditions (equations) linearly.

For Example:

The third term of an A.P. is 7 and the seventh term exceeds three times the third term

by 2. Find the first term, the common difference and the sum of first 20 terms.

a3 = 7

a7 = 3a3 + 2

a + 2d = 7 ..... (i)

a + 6d = 3 × 7 + 2

a + 6d = 23 ..... (ii)

Solving (i) and (ii)

4d = 16

d= 4

Putting the value of d in equation (i),

a+2×4= 7
a= –1

 S20 = 20 [(2  (−1) + (20 − 1)  4]
2

= 10 [–2 + 76]

S20 = 10 (74)

So, S20 = 740

d= 4
a = –1

3. First negative term
To find the first negative term of the given sequence, put an  0 .

For example:

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To find the first negative term of the sequence 20, 19 1 ,18 1 ,17 3 ,.... .
424

Here a = 20 and d = − 3
4

Then, an  0
 a + (n − 1)d  0
 20 + (n − 1)   − 3   0

 4
 83 − 3n  0

44
 83 − 3n  0
 3n  83  n  27 2  n  28

3
Thus, 28th term of the given sequence is the first negative term.

4. Finding term from the end
To find the nth term from the end, apply an = l + (n − 1)(−d ) or an = l − (n − 1)d , where
l is the last term.
or take last term (l) of the given sequence as ‘a’ and take common difference as (−d)
For example:
Find the 5th term from the end of the AP, 17, 14, 11, ….., −40
1st method
l = −40,d = 14 −17 = −3

Using l − (n −1)d

5th term from the end will be
= −40 − (5 −1) −3 = −40 − 4  −3

= −40 + 12 = −28
2nd method
Sequence can be written as −40, −37, …. 11, 14, 17
 a = −40

d = −37 − (−40) = −37 + 40 = 3

n=5
Using an = a + (n − 1)d

= − 40 + (5 − 1)  3 = −40 + 4  3

= − 40 + 12 = −28

5. Condition for terms to be in A.P.
If three terms a, b, c are in A.P. then b − a = c − b, i.e., 2b = a + c.
For example:
If 2x, x + 10,3x + 2 are in A.P., find the value of x.

Since, 2x,x +10,3x + 2 are in A.P.

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 2(x + 10) = 2x + (3x + 2)

 2x + 20 = 5x + 2
 3x = 18
 x=6

6. Choice of terms in A.P.

No. of terms Terms Common difference
d
3 a − d, a, a + d 2d
d
4 a − 3d, a − d, a + d, a + 3d
2d
5 a − 2d, a − d, a, a + d, a + 2d

a − 5d, a − 3d, a − d, a + d,
6 a + 3d, a + 5d

For example:
The sum of three numbers in A.P. is -3, and their product is 8. Find the numbers.

Let the number be (a − d ),a,(a + d ) . Then

Sum = −3  (a − d )+ a + (a + d ) = −3

 3a = −3 …(i)
[from (i)]
 a = −1

Now, Product = 8

 (a − d )(a)(a + d ) = 8

( ) a a2 − d 2 = 8
( ) (− 1)1− d 2 = 8

 d 2 = 9  d = 3
If d = 3 the numbers are − 4,−1, 2. If d = -3, the numbers are 2, -1, -4.

Thus, the numbers are − 4,−1, 2 or 2,−1,−4.
7. Sum of first n-terms of an A.P.

Sn = n [2a + (n − 1)d ]
2

where a = first term, d = common difference.

8. Sn is also given by the expression

Sn = n (a + l ),
2

where l = last term as l = a + (n − 1)d = an .

For example:

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

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Here, a = 5, l = an = 45, Sn = 400

 Sn = n a + l
2

 400 = n 5 + 45 = 25n

2

 n = 400 = 16
25

Also l = 45

 an = a + (n −1)d = 45
 5 + (16 −1) d = 45 [ a = 5, n = 16]

15d = 45 − 5 = 40

 d = 40 = 8
15 3

Hence, number of terms(n) = 16, d = 8
3

9. If sum (Sn ) of n terms of a sequence is given then nth term (an ) of the sequence
can be determined by nth term an = Sn − Sn −1 and common difference
d = an − an −1 = Sn − 2Sn −1 + Sn −2 .
For example:
If the sum of first n terms of an A.P. is 3n2 − 2n, find the A.P. and its 19th term.

Sn = 3n2 − 2n
Put n = 1 ; S1 = 3 − 2 = 1
Put n = 2 ; S2 = 3(2)2 − 2(2)

=3×4–48
Put n = 3 ; S3 = 3(3)2 − 2(3)

= 27 – 6 = 21
 a = S1 = 1

a2 = S2 − S1 = 7
a3 = S3 − S2 = 21− 8 = 13
 Required A.P. is; 1, 7, 13, ..........
a = 1, d = 6
a19 = a + 18d
= 1 + 18 × 6
= 1 + 108 = 109

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