the-sum-of-the-first-three-terms-of-an-AP-is-21-and-the-sum-of-the-first-five-terms-is-55-find-1-the-first-term-2-common-difference-3-the-sum-of-the-first-ten-term-of-the-sequence-

FacebookTweetPin

Question Number 188758 by Humble last updated on 06/Mar/23

$$thesumofthefirstthreeterms$$$$ofanAPis21andthesumofthe$$$$firstfivetermsis55.find.$$$$\left(1\right)thefirstterm$$$$\left(2\right)commondifference$$$$\left(3\right)thesumofthefirsttenterm$$$$ofthesequence$$$$$$

Answered by floor(10²Eta[1]) last updated on 06/Mar/23

$${\mathrm{S}}_3=21,{\mathrm{S}}_5=55$$$$\frac{({\mathrm{a}}_1+{\mathrm{a}}_3)3}{2}=21,\frac{({\mathrm{a}}_1+{\mathrm{a}}_5)5}{2}=55$$$$\Rightarrow {\mathrm{a}}_1+{\mathrm{a}}_3=14,{\mathrm{a}}_1+{\mathrm{a}}_5=22$$$$\Rightarrow {\mathrm{a}}_1+({\mathrm{a}}_1+2\mathrm{r})=14,{\mathrm{a}}_1+({\mathrm{a}}_1+4\mathrm{r})=22$$$$\Rightarrow {\mathrm{a}}_1+\mathrm{r}=7,{\mathrm{a}}_1+2\mathrm{r}=11$$$$\Rightarrow \mathrm{r}=4,{\mathrm{a}}_1=3$$$$\left(1\right):3$$$$\left(2\right):4$$$$\left(3\right):{\mathrm{S}}_{10}=\frac{({\mathrm{a}}_1+{\mathrm{a}}_{10})10}{2}=210$$

Answered by Rasheed.Sindhi last updated on 06/Mar/23

$$LettisthirdtermoftheAP$$$$(t-2d)+(t-d)+t+(t+d)+(t+2d)=55$$$$5t=55\Rightarrow t=11$$$$(t-2d)+(t-d)+t=21$$$$3t-3d=21$$$$3\left(11\right)-3d=21\Rightarrow d=\frac{33-21}{3}=4$$$$firstterm=a=t-2d=11-2\left(4\right)=3$$$$S_{10}=\frac{10}{2}[2\left(3\right)+(10-1)\left(4\right)]$$$$=5(6+36)=210$$

Answered by Rasheed.Sindhi last updated on 06/Mar/23

$$Letu\mathrm{\&}tare2nd\mathrm{\&}3rdtermsrespectively.$$$$discommondifference.$$$$\bullet (u-d)+u+(u+d)=21$$$$3u=21\Rightarrow u=7\left(2ndterm\right)$$$$\bullet (t-2d)+(t-d)+t+(t+d)+(t+2d)=55$$$$5t=55\Rightarrow t=11\left(3rdterm\right)$$$$\bullet d=t-u=11-7=4$$$$\bullet Firstterm=u-d=7-4=3$$$$\bullet S_{10}=\frac{10}{2}[2\left(3\right)+(10-1)\left(4\right)]$$$$=5(6+36)=5\times 42=210$$

Commented by Humble last updated on 07/Mar/23

$$thanks,sir.$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin