NCERT/CBSE MATHS Class 10 Ex-5.2 Q No.11 Solutions : Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?
Hy Friends Welcome On NCERT MATHS SOLUTIONS !! Today we are going to solve the Question : ”Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?” of Exercise 5.2 Question No 11 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. 11 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 – 11 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. 11 :
Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?
Given A.P. is 3, 15, 27, 39, …
a = 3
d = a2 − a1 = 15 − 3 = 12
a54 = a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636 = 639
132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let nth term be 771.
an = a + (n − 1) d
771 = 3 + (n − 1) 12
768 = (n − 1) 12
(n − 1) = 64
n = 65
Therefore, 65th term was 132 more than 54th term.
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