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

Question 27

Step 1: Recall formulas.

• General term of an AP: \(a_n = a + (n-1)d\)
• Sum of first \(n\) terms: \(S_n = \dfrac{n}{2}[2a + (n-1)d]\)

Step 2: Write equations using given terms.

We know:

• \(a_4 = a + 3d = -15\)  ...(i)
• \(a_9 = a + 8d = -30\)  ...(ii)

Step 3: Subtract the equations to find \(d\).

(ii) – (i):

\((a + 8d) - (a + 3d) = -30 - (-15)\)

\(5d = -15\)

So, \(d = -3\).

Step 4: Substitute \(d\) in equation (i) to find \(a\).

\(a + 3(-3) = -15\)

\(a - 9 = -15\)

\(a = -6\).

Step 5: Use the sum formula for \(n = 17\).

\(S_{17} = \dfrac{17}{2}[2a + (17-1)d]\)

\(= \dfrac{17}{2}[2(-6) + 16(-3)]\)

\(= \dfrac{17}{2}[-12 - 48]\)

\(= \dfrac{17}{2}[-60]\)

\(= 17 \times -30\)

\(= -510\).

Final Answer: \(S_{17} = -510\)