NCERT/CBSE MATHS Class 10 Ex-5.2 Q No.8 Solutions : An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Hy Friends Welcome On NCERT MATHS SOLUTIONS !! Today we are going to solve the Question : ”An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.” of Exercise 5.2 Question No 8 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. 8 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 – 8 Solutions . Hope you like this post about Class 10th Maths Solution.
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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
Exercise- 5.2 Question No. 8 :
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
a3 = 12
a50 = 106
We know that,
an = a + (n − 1) d
a3 = a + (3 − 1) d
12 = a + 2d … (i)
Similarly, a50 = a + (50 − 1) d
106 = a + 49d … (ii)
On subtracting (i) from (ii), we get
94 = 47d
d = 2
From equation (i), we get
12 = a + 2 (2)
a = 12 − 4 = 8
a29 = a + (29 − 1) d
a29 = 8 + (28)2
a29 = 8 + 56 = 64
Therefore, 29th term is 64.
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