Question

Which of the following list of numbers forms an AP?
(i) 4, 4 + $\sqrt{3}$ , 4 + 2$\sqrt{3}$, 4 + 3$\sqrt{3}$, 4 + 4$\sqrt{3}$

(ii) 0.3, 0.33, 0.333, 0.3333, 0.33333

(iii) $\frac{3}{5}$, $\frac{6}{5}$ , $\frac{9}{5}$ , $\frac{12}{5}$ , 3

(iv) -$\frac{1}{5}$ , -$\frac{1}{5}$ , -$\frac{1}{5}$ , -$\frac{1}{5}$ , -$\frac{1}{5}$

• A. Only (i) and (ii)
• B. Only (ii) and (iii)
• C. Only (iii) and (iv)
• D. Only (i), (iii) and (iv)

Answer 1

The correct option is D

Only (i), (iii) and (iv)

Consider each series
(i) 4, 4 + $\sqrt{3}$, 4 + 2 $\sqrt{3}$, 4 + 3 $\sqrt{3}$, 4 + 4 $\sqrt{3}$

Difference between first two consecutive terms = 4 + $\sqrt{3}$ - 4 = $\sqrt{3}$

Difference between third and second consecutive terms = 4 + 2 $\sqrt{3}$ - 4 + $\sqrt{3}$ = $\sqrt{3}$

Difference between fourth and third consecutive terms = 4 + 3 $\sqrt{3}$– 4 + 2 $\sqrt{3}$ = $\sqrt{3}$

Since $a_2–a_1=a_3-a_2=a_4-a_3$

This series is an AP

(ii) 0.3, 0.33, 0.333, 0.3333, 0.33333

Difference between first two consecutive terms = 0.33 - 0.3 = 0.03

Difference between third and second consecutive terms = 0.333 - 0.33 = 0.003

Since $a_2–a_1\ne a_3-a_2$
This series is not an AP

(iii) $\frac{3}{5}$, $\frac{6}{5}$, $\frac{9}{5}$, $\frac{12}{5}$, 3

Difference between first two consecutive terms = $\frac{6}{5}–\frac{3}{5}$ = $\frac{3}{5}$

Difference between third and second consecutive terms = $\frac{9}{5}–\frac{6}{5}$ = $\frac{3}{5}$

Difference between fourth and third consecutive terms = $\frac{12}{5}$ – $\frac{9}{5}$ = $\frac{3}{5}$

Since $a_2–a_1=a_3–a_2=a_4–a_3$
This series is an AP

(iv) -$\frac{1}{5}$, -$\frac{1}{5}$, -$\frac{1}{5}$, -$\frac{1}{5}$, -$\frac{1}{5}$

Difference between first two consecutive terms = -$\frac{1}{5}$ – (- $\frac{1}{5}$) = 0

Difference between third and second consecutive terms = -$\frac{1}{5}$ – (-$\frac{1}{5}$) = 0

Difference between fourth and third consecutive terms = -$\frac{1}{5}$ – (-$\frac{1}{5}$) = 0

Since $a_2–a_1=a_3–a_2=a_4–a_3$
This series is an AP