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Class 10Algebra2:34Published 25 Feb 2024

Important Questions on Arithmetic Progression (Part 1)

Three worked exam questions on arithmetic progressions for Class 10, covering how to read the common difference from the nth term and how to use the sum formula.

This short lesson walks through three important Class 10 questions on arithmetic progressions. You will see how to find the common difference straight from the nth term, how to recover terms from the sum of the first n terms, and how to count how many terms an AP has. Each example is solved step by step so the method is clear.

What you'll learn

  • How to read the common difference directly from the nth term formula
  • How to find individual terms from the sum of the first n terms
  • How to count the number of terms in an arithmetic progression

Lesson chapters

0:00Common difference from the nth term
0:39Finding terms from the sum formula
1:34Counting the number of terms
2:19Note: the first term equals the first sum

Lesson notes

This lesson works through three important Class 10 questions on arithmetic progressions (AP). Each one is solved step by step.

Question 1: Common difference from the nth term

We are given the nth term of an AP as

an=3n+5a_n = 3n + 5

and asked to find the common difference dd.

When the nth term is written as a linear expression in nn, the common difference is simply the coefficient of nn. Here that coefficient is 33, so

d=3d = 3

Question 2: Finding terms from the sum formula

We are given the sum of the first nn terms as

Sn=n2S_n = n^2

and asked to find the common difference dd.

First sum and first term. Put n=1n = 1:

S1=12=1,a1=S1=1S_1 = 1^2 = 1, \qquad a_1 = S_1 = 1

Second sum. Put n=2n = 2:

S2=22=4S_2 = 2^2 = 4

Second term. The second term is the second sum minus the first sum:

a2=S2S1=41=3a_2 = S_2 - S_1 = 4 - 1 = 3

Common difference.

d=a2a1=31=2d = a_2 - a_1 = 3 - 1 = 2

Question 3: Counting the number of terms

We are given

an=500,a1=50,d=5a_n = 500, \qquad a_1 = 50, \qquad d = 5

and asked to find the number of terms nn. Rearranging the nth term formula gives

n=ana1d+1n = \frac{a_n - a_1}{d} + 1

Substituting the values:

n=500505+1=4505+1=90+1=91n = \frac{500 - 50}{5} + 1 = \frac{450}{5} + 1 = 90 + 1 = 91

So the AP has 9191 terms.

Key takeaways

  • If an=pn+qa_n = pn + q, the common difference is the coefficient pp.
  • The first term equals the first sum, a1=S1a_1 = S_1, and any term is an=SnSn1a_n = S_n - S_{n-1}.
  • The number of terms is n=ana1d+1n = \dfrac{a_n - a_1}{d} + 1.