MCQ
Which of the following is not an $A.P.?$
- A$-1.2,0.8,2.8,\ \dots$
- B$3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},\ \dots$
- ✓$\frac43,\frac73,\frac93,\frac{12}3,\ \dots$
- D$\frac{-1}{5},\frac{-2}{5},\frac{-3}{5},\ \dots$
✓
Answer
Correct option: C.
$\frac43,\frac73,\frac93,\frac{12}3,\ \dots$
For Arithmetic progression the series have common difference in two consecutive terms i.e. $\text{a}_2 - \text{a}_1 = \text{a}_3 - \text{a}_2$
$0.8-(-1.2)=2.8-0.8$
$2=2$
So, the series given is an $A.P$.
$3+\sqrt{2}-3=3+3\sqrt{2}-3+2\sqrt{2}$
$\sqrt{2}=\sqrt{2}$
So, the series given is an $A.P$.
$\frac73,\frac{-4}3,\frac{-9}{3},\frac{-7}3$
$\frac33=\frac23$
$3\neq2$
So, the series given is not an $A.P$.
$\frac{-2}5-\Big(\frac{-1}{5}\Big)-\Big(\frac{-3}{5}\Big)-\Big(\frac{-2}{5}\Big)$
$\frac{-2}{5}+\frac15=\frac{-3}5+\frac25$
$\frac{-2+1}{5}=\frac{-3+2}{5}$
$-1=-1$
So, the series given is an $A.P.$
Therefore, option $ C$ is correct.
- $-1.2,0.8,2.8,\ \dots$
$0.8-(-1.2)=2.8-0.8$
$2=2$
So, the series given is an $A.P$.
- $3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},\ \dots$
$3+\sqrt{2}-3=3+3\sqrt{2}-3+2\sqrt{2}$
$\sqrt{2}=\sqrt{2}$
So, the series given is an $A.P$.
- $\frac43,\frac73,\frac93,\frac{12}3,\ \dots$
$\frac73,\frac{-4}3,\frac{-9}{3},\frac{-7}3$
$\frac33=\frac23$
$3\neq2$
So, the series given is not an $A.P$.
- $\frac{-1}{5},\frac{-2}{5},\frac{-3}{5},\ \dots$
$\frac{-2}5-\Big(\frac{-1}{5}\Big)-\Big(\frac{-3}{5}\Big)-\Big(\frac{-2}{5}\Big)$
$\frac{-2}{5}+\frac15=\frac{-3}5+\frac25$
$\frac{-2+1}{5}=\frac{-3+2}{5}$
$-1=-1$
So, the series given is an $A.P.$
Therefore, option $ C$ is correct.
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