NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.1

Question 17

We are given:

• First term, \(a = 1\)
• Last term, \(a_n = 20\)
• Sum of \(n\) terms, \(S_n = 399\)

Step 1: Recall the formula for the sum of \(n\) terms of an AP:

\[ S_n = \dfrac{n}{2} (a + a_n) \]

Step 2: Substitute the given values:

\[ 399 = \dfrac{n}{2} (1 + 20) \]

Step 3: Simplify inside the bracket:

\[ 399 = \dfrac{n}{2} (21) \]

Step 4: Multiply:

\[ 399 = \dfrac{21n}{2} \]

Step 5: Remove the fraction by multiplying both sides by 2:

\[ 798 = 21n \]

Step 6: Divide both sides by 21:

\[ n = \dfrac{798}{21} = 38 \]

Therefore, \(n = 38\).