Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=n(n+2)$

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

$a_{n}=n(n+2)$

Substituting $n=1,2,3,4$ and $5,$ we obtain

$a_{1}=1(1+2)=3$

$a_{2}=2(2+2)=8$

$a_{3}=3(3+2)=15$

$a_{4}=4(4+2)=24$

$a_{5}=5(5+2)=35$

Therefore, the required terms are $3,8,15,24$ and $35 .$

Similar Questions

Let the coefficients of the middle terms in the expansion of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}$ and $\left(1-\frac{\beta}{2} x\right)^{6}, \beta>0$, respectively form the first three terms of an $A.P.$ If $d$ is the common difference of this $A.P.$, then $50-\frac{2 d}{\beta^{2}}$ is equal to.

  • [JEE MAIN 2022]

If $a _{1}, a _{2}, a _{3} \ldots$ and $b _{1}, b _{2}, b _{3} \ldots$ are $A.P.$ and $a_{1}=2, a_{10}=3, a_{1} b_{1}=1=a_{10} b_{10}$ then $a_{4} b_{4}$ is equal to

  • [JEE MAIN 2022]

If $3^{2 \sin 2 \alpha-1},14$ and $3^{4-2 \sin 2 \alpha}$ are the first three terms of an $A.P.$ for some $\alpha$, then the sixth term of this $A.P.$ is 

  • [JEE MAIN 2020]

Find the $17^{\text {th }}$ and $24^{\text {th }}$ term in the following sequence whose $n^{\text {th }}$ term is $a_{n}=4 n-3$

If ${a^2},\,{b^2},\,{c^2}$ be in $A.P.$, then $\frac{a}{{b + c}},\,\frac{b}{{c + a}},\,\frac{c}{{a + b}}$ will be in