Arithmetic Progressions Class 10 CBSE

Download NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.3 – Arithmetic Progressions

1. Find the sum of the following APs.

(i) 2, 7, 12, …. to 10 terms.

Solution:

For this A.P.
a = 2
d = 7-2 = 5
Number of term (n) = 10
As we know that,
S_n = [2a+(n-1)d]
S₁₀ = [2×2+(10-1)5]
S₁₀ = 5[4+45]
S₁₀ = 245

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Download NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 – Arithmetic Progressions

1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for the each additional km.

Solution: Taxi fare for 1^st 1 km = ₹15.
Taxi fare for first 2 km = ₹15 + ₹8 = ₹23
Taxi fare for first 3 km = ₹23+₹8 = ₹31
Taxi fare for first 4 km = ₹31+₹8 = ₹39
Therefore, we observe that, 15, 23, 31, 39….. forms an A.P. because the common difference for each km is same i.e. 8.
Hence, the given situation makes an arithmetic progression.

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Download NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.2 – Arithmetic Progressions

A. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the n^th term of the AP:

Solution : If a be the first term and d be the common difference of an AP then the n^th term of that AP is given by

a_n = a+(n-1)d

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