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# NCERT/CBSE MATHS Class 10 Ex-5.2 Q No.3 Solutions : In the following APs find the missing term in the boxes.

Hy Friends Welcome On NCERT MATHS SOLUTIONS !! Today we are going to solve the Question : In the following APs find the missing term in the boxes.of Exercise 5.2 Question No 3 Solutions (Arithmetic Progressions) of Class 10th, which will prove to be very helpful for you.

# CBSE/NCERT MATHS Class 10 Ex-5.2 Q No. 3 Solutions

If you are a student of CBSE, today we are going to give you the CBSE / NCERT Chapter : Arithmetic Progressions Exercise 5.2 Question No – 3 Solutions . Hope you like this post about Class 10th Maths Solution.

Here we would like to tell you that there are other websites on which the students demanded the Solution of NCERT MATHS Chapters, we had promised them that we will soon send you the questions of Class 9, And we completed our promise to them.We asked the good knowledgeable teachers of mathematics and asked them to solve the issues of CBSE’s mathematical questions and also said that the answers to our questions are such that the students have no difficulty in solving all those questions.

### Arithmetic Progressions CHAPTER : 5

n this progression, for a given series, the terms used are the first term, the common difference between the two terms and nth term.

Suppose, a1, a2, a3, ……………., an is an AP, then; the common difference “ d ” can be obtained as; In this progression, for a given series, the terms used are the first term, the common difference between the two terms and nth term.

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

### Class 10 Exercise 5.2 Question No. 3 Solutions

(i) 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) 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 get
– 10 = 2d
d = – 5
From equation (i), we get
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.,
= 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 = 3d3d = 9/2
d = 3/2

a2 = a + (2 – 1) d
a2 = + 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) For this A.P.,
a = −4 and
a6 = 6
We know that,
an = a + (n − 1) d
a6 = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d = 2
a2 = a + d = − 4 + 2 = −2
a3 = a + 2d = − 4 + 2 (2) = 0
a4 = a + 3d = − 4 + 3 (2) = 2
a5 = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.

(v) For this A.P.,
a2 = 38
a6 = −22
We know that
an = a + (n − 1) d
a2 = a + (2 − 1) d
38 = a + d … (i)
a6 = a + (6 − 1) d
−22 = a + 5d … (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a2 − d = 38 − (−15) = 53
a3 = + 2= 53 + 2 (−15) = 23
a4 = a + 3d = 53 + 3 (−15) = 8
a5 = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.