Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 74284 by arthur.kangdani@gmail.com last updated on 21/Nov/19

Commented by mr W last updated on 21/Nov/19

$$a_2=4$$$$a_{21}=99$$$$\mathrm{\Sigma }=\frac{(4+99)\times 20}{2}=1030$$

Commented by TawaTawa last updated on 21/Nov/19

$$\mathrm{when}\mathrm{n}=2$$$$5\left(2\right)-6=10-6=4$$$$\mathrm{when}\mathrm{n}=3$$$$5\left(3\right)-6=15-6=9$$$$\mathrm{when}\mathrm{n}=4$$$$5\left(4\right)-6=20-6=14$$$$.$$$$.$$$$.$$$$\mathrm{When}\mathrm{n}=21$$$$5\left(21\right)-6=105-6=99$$$$\mathrm{So},\mathrm{AP}:$$$$4+9+14+...+99$$$$\mathrm{a}=4,\mathrm{d}=5,\mathrm{L}=99,\mathrm{n}=20\left(\mathrm{from}2\mathrm{to}21\right)$$$$\mathrm{Now},$$$$\therefore {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{L})$$$$\therefore {\mathrm{S}}_{20}=\frac{20}{2}(4+99)$$$$\therefore {\mathrm{S}}_{20}=10.\left(103\right)$$$$\therefore {\mathrm{S}}_{20}=1030$$$$\therefore \underset{\mathrm{n}=2}{\sum ^{21}}(5\mathrm{n}-6)=1030$$

Commented by TawaTawa last updated on 21/Nov/19

$$\mathrm{Wow},\mathrm{i}\mathrm{got}\mathrm{it}.\mathrm{sir}\mathrm{mrW}\mathrm{thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Commented by mathmax by abdo last updated on 21/Nov/19

$$letS=\sum _{n=2}^{21}(5n-6)changementofindicen-1=kgive$$$$S=\sum _{k=1}^{20}(5(k+1)-6)=\sum _{k=1}^{20}(5k-1)thesequenc{u}_k=5k-1is$$$$srithmeticwithr=5and{u}_1=4\Rightarrow$$$$S=u_1+u_2+...+u_{20}=\frac{20}{2}(u_1+u_{20})=10(4+99)=10\times 103=1030.$$

Answered by $@ty@m123 last updated on 21/Nov/19

$$=\left\{\underset{n=1}{\sum ^{21}}(5n-6)\right\}-(5\times 1-6)$$$$=\{5\mathrm{\Sigma }n-\mathrm{\Sigma }6\}-(-1)$$$$=5\times \frac{n(n+1)}{2}-6n+1$$$$=5\times \frac{21\times 22}{2}-6\times 21+1$$$$=1155-126+1$$$$=1030$$

Answered by malwaan last updated on 22/Nov/19

$$<4;9;14;19;....;94;99>$$$$\mathit{n}=20;{\mathit{u}}_1=4;\mathit{r}=5$$$$\Rightarrow \mathit{S}=\frac{\mathit{n}}{2}[2{\mathit{u}}_1+(\mathit{n}-1)\mathit{r}]$$$$=\frac{20}{2}[2\times 4+(20-1)\times 5]$$$$=10(8+95)=10\times 103=1030$$

Commented by $@ty@m123 last updated on 22/Nov/19

$$Whenlasttermisknown,$$$$itiseasiertousefollowingformula:$$$$\mathit{S}=\frac{\mathit{n}}{2}(\mathit{a}+\mathit{l})$$$$wheren=no.ofterms$$$$a=firstterm$$$$l=lastterm$$$$\Rightarrow \mathit{S}=\frac{20}{2}(4+99)=1030$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com