Write the first five terms of the following sequence and obtain the corresponding series :

$a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n\,>\,2$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

$a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n\,>\,2$

$\Rightarrow a_{3}=a_{2}-1=2-1=1$

$a_{4}=a_{3}-1=1-1=0$

$a_{5}=a_{4}-1=0-1=-1$

Hence, the first five terms of the sequence are $2,2,1,0$ and $-1$ 

The corresponding series is $2+2+1+0(-1)+\ldots$

Similar Questions

A man deposited $Rs$ $10000$ in a bank at the rate of $5 \%$ simple interest annually. Find the amount in $15^{\text {th }}$ year since he deposited the amount and also calculate the total amount after $20$ years.

If $a, b, c, d$ are in $G.P.,$ prove that $\left(a^{n}+b^{n}\right),\left(b^{n}+c^{n}\right),\left(c^{n}+d^{n}\right)$ are in $G.P.$

Let $x_n, y_n, z_n, w_n$ denotes $n^{th}$ terms of four different arithmatic progressions with positive terms. If $x_4 + y_4 + z_4 + w_4 = 8$ and $x_{10} + y_{10} + z_{10} + w_{10} = 20,$ then maximum value of $x_{20}.y_{20}.z_{20}.w_{20}$ is-

If the sum of a certain number of terms of the $A.P.$ $25,22,19, \ldots \ldots .$ is $116$ Find the last term

For $\mathrm{x} \geq 0$, the least value of $\mathrm{K}$, for which $4^{1+\mathrm{x}}+4^{1-\mathrm{x}}$, $\frac{\mathrm{K}}{2}, 16^{\mathrm{x}}+16^{-\mathrm{x}}$ are three consecutive terms of an $A.P.$ is equal to :

  • [JEE MAIN 2024]