CBSE Class 10 Mathematics Standard 2023: question 26(B)

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. For the A.P. 65, 61, 57, 53, …: $a=65$ and $d=61-65=-4$.

   ½ mark

2. Let $a_n$ be the first negative term. Then $a_n<0$:

   $a+(n-1)d<0$

   $65+(n-1)(-4)<0$

   $69-4n<0$

   1 mark

3. $4n>69$, so $n>\frac{69}{4}=17\frac{1}{4}$

   1 mark

4. The least whole number greater than $17\frac{1}{4}$ is $n=18$.

   Check: $a_{17}=65-64=1>0$ and $a_{18}=65-68=-3<0$.

   So the 18th term is the first negative term.

   ½ mark

Answer: 18th term (which equals $-3$)

Revise the chapter

• Chapter: Arithmetic Progressions (notes, formulas, where marks are lost and every past question)