NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.3

Question 32

Step 1: Write down what is given.

• First term, \(a = -15\)
• Common difference, \(d = -13 - (-15) = 2\)
• Required sum, \(S_n = -55\)

Step 2: Recall the formula for the sum of first \(n\) terms of an AP:

\[ S_n = \dfrac{n}{2} \big( 2a + (n-1)d \big) \]

Step 3: Substitute the values \(a = -15\) and \(d = 2\):

\[ S_n = \dfrac{n}{2} \big( 2(-15) + (n-1) \times 2 \big) \]

\[ S_n = \dfrac{n}{2} ( -30 + 2n - 2 ) \]

\[ S_n = \dfrac{n}{2} ( 2n - 32 ) \]

\[ S_n = n(n - 16) \]

Step 4: We are told that \(S_n = -55\). So,

\[ n(n - 16) = -55 \]

\[ n^2 - 16n + 55 = 0 \]

Step 5: Solve the quadratic equation.

Factorize: \[ n^2 - 16n + 55 = 0 \]

\[ (n - 5)(n - 11) = 0 \]

So, \(n = 5\) or \(n = 11\).

Step 6: Check both answers.

• For \(n = 5\): First 5 terms are \(-15, -13, -11, -9, -7\). Their sum is \(-55\). ✔
• For \(n = 11\): First 11 terms add up to \(-55\) again because the extra 6 terms after the 5th can be grouped in pairs like \((-5 + 5), (-3 + 3), (-1 + 1)\), each pair giving 0. So the total remains \(-55\). ✔

Final Answer: The required number of terms is 5 or 11.