Question

Which of the following series forms an AP?

(i) a – 2d, a – d, a, a + d, a + 2d

(ii)a, a + d, a + 2d, a + 3d, a + 4d

(iii) a – 3d, a – d, , a + d, a + 3d

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

Answer 1

The correct option is D

All (i),(ii) and (iii)

Consider each series(i) a – 2d, a – d, a, a + d, a + 2d

Difference between first two consecutive terms = a – d –(a – 2d) = d

Difference between third and second consecutive terms = a – (a – d) = d

Difference between fourth and third consecutive terms = a + 2d – (a + d) = d

Since $a_2$ – $a_1$ = $a_3$ – $a_2$= $a_4$ – $a_3$

This series is an AP

Consider each series(ii) a, a + d, a + 2d, a + 3d, a + 4d

Difference between first two consecutive terms = a + d –a = d

Difference between third and second consecutive terms = a + 2d - (a + d) = d

Difference between fourth and third consecutive terms = a + 3d – (a +2d) = d

Since $a_2$ – $a_1$ = $a_3$ – $a_2$ = $a_4$ – $a_3$

This series is an AP

(iii) a – 3d, a – d, a + d, a + 3d

Difference between first two consecutive terms = a – d – (a – 3d) = 2d

Difference between third and second consecutive terms = a + d – (a – d) = 2d

Difference between fourth and third consecutive terms = a +3d – (a +d) = 2d

Since $a_2$ – $a_1$ = $a_3$ – $a_2$ = $a_4$ – $a_3$

This series is an AP