Find the next term in the following sequence: a-2d, a-d, a, a+d, ...
(A) a + 2d
(B) a - 2d
(C) 4a + 2d
(D) 2a + 2d
$[1]$

Open in App

Solution

Option A is the correct answer.

let $A_{1}$ and D be the first term and the common difference respectively.
$A_{1} = a - 2 d$
D=a-d-a+2d=d
$A_{5} = A_{1} + 4 D = a - 2 d + 4 d = a + 2 d$

Suggest Corrections

Similar questions

$a + (a + d) + (a + 2 d) + . . . . + (a + (n - 1) d) = \frac{n}{2} [2 a + (n - 1) d]$

Find the next term in the following sequence: a-2d, a-d, a, a+d,...

{(\frac{2^{a}}{2^{b}})}^{(a + b)} . {(\frac{2^{b}}{2^{c}})}^{(b + c)} . {(\frac{2^{c}}{2^{d}})}^{(c + d)} . {(\frac{2^{d}}{2^{a}})}^{(d + a)} = 1

a > 0, d > 0

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{a} & \frac{1}{a (a + d)} & \frac{1}{(a + d) (a + 2 d)} \frac{1}{a (a + d)} & \frac{1}{(a + d) (a + 2 d)} & \frac{1}{(a + 2 d) (a + 3 d)} \frac{1}{(a + 2 d)} & \frac{1}{(a + 2 d) (a + 3 d)} & \frac{1}{(a + 3 d) (a + 4 d)} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{a} & \frac{1}{a (a + d)} & \frac{1}{(a + d) (a + 2 d)} \frac{1}{(a + d)} & \frac{1}{(a + d) (a + 2 d)} & \frac{1}{(a + 2 d) (a + 3 d)} \frac{1}{(a + 2 d)} & \frac{1}{(a + 2 d) (a + 3 d)} & \frac{1}{(a + 3 d) (a + 4 d)} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣

Join BYJU'S Learning Program