NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.2

Question 1

Key formula for an AP: The n-th term of an AP is given by

\[a_n = a + (n - 1)d\]

where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the term number.

(i) Row (i): \(a = 7,\ d = 3,\ n = 8\)

Use \(a_n = a + (n - 1)d\):

\[a_8 = 7 + (8 - 1) \cdot 3 = 7 + 7 \cdot 3 = 7 + 21 = 28\]

So, \(a_n = 28\).

(ii) Row (ii): \(a = -18,\ n = 10,\ a_{10} = 0\)

Use \(a_n = a + (n - 1)d\):

\[0 = -18 + (10 - 1)d = -18 + 9d\]

\[9d = 18 \Rightarrow d = \dfrac{18}{9} = 2\]

So, \(d = 2\).

(iii) Row (iii): \(d = -3,\ n = 18,\ a_{18} = -5\)

Again, \(a_n = a + (n - 1)d\):

\[-5 = a + (18 - 1)(-3) = a + 17(-3) = a - 51\]

\[a - 51 = -5 \Rightarrow a = -5 + 51 = 46\]

So, \(a = 46\).

(iv) Row (iv): \(a = -18.9,\ d = 2.5,\ a_n = 3.6\)

Use \(a_n = a + (n - 1)d\):

\[3.6 = -18.9 + (n - 1) \cdot 2.5\]

Move \(-18.9\) to the left:

\[3.6 + 18.9 = (n - 1) \cdot 2.5\]

\[22.5 = (n - 1) \cdot 2.5\]

\[n - 1 = \dfrac{22.5}{2.5} = 9 \Rightarrow n = 10\]

So, \(n = 10\).

(v) Row (v): \(a = 3.5,\ d = 0,\ n = 105\)

If \(d = 0\), every term of the AP is equal to the first term.

So,

\[a_n = a = 3.5\]

Thus, \(a_n = 3.5\).

Summary: Each unknown in the table is found by rearranging and applying the basic n-th term formula of an AP.