Question

Which of the following series forms an AP?

$\left(i\right)a–2d,a–d,a,a+d,a+2d$

$\left(ii\right)a,a+d,a+2d,a+3d,a+4d$

$\left(iii\right)a–3d,a–d,a+d,a+3d$

• A. $\left(ii\right)and\left(iii\right)$
• B. $\left(i\right)and\left(ii\right)$
• C. $\left(i\right),\left(ii\right)and\left(iii\right)$
• D. $\left(i\right)and\left(iii\right)$

Answer 1

The correct option is C

$\left(i\right),\left(ii\right)and\left(iii\right)$

Consider each series $\left(i\right)a–2d,a–d,a,a+d,a+2d$

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 $a_2–a_1=a_3–a_2=a_4–a_3$
So, the sequence is in AP
Similarly, $\left(ii\right)$ has common diffrence of $d$ and $\left(iii\right)$ has commonn difference of $2d$ and all are in A.P.