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What-is-the-sum-of-first-3n-term-of-an-AP-if-the-sunm-of-first-n-term-is-2n-and-sum-of-first-2n-term-is-5n-




Question Number 61137 by Tawa1 last updated on 29/May/19
What is the sum of first 3n term of an AP , if the sunm of first n term is  2n  and sum of first 2n term is  5n
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{3n}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:,\:\mathrm{if}\:\mathrm{the}\:\mathrm{sunm}\:\mathrm{of}\:\mathrm{first}\:\mathrm{n}\:\mathrm{term}\:\mathrm{is} \\ $$$$\mathrm{2n}\:\:\mathrm{and}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{2n}\:\mathrm{term}\:\mathrm{is}\:\:\mathrm{5n} \\ $$
Answered by Kunal12588 last updated on 29/May/19
S_n =2n  ⇒(n/2)(2a+(n−1)d)=2n  ⇒2a+nd−d−4=0  S_(2n) =5n  ⇒((2n)/2)(2a+(2n−1)d)=5n  ⇒2a+2nd−d−5=0    subtracting the red ones  nd−1=0  ⇒nd=1    putting value of nd in either of red lines  2a−d−3=0  ⇒2a−d=3    S_(3n) =((3n)/2)(2a+(3n−1)d)  =((3n)/2)(2a−d+3nd)  using blue ones  S_(3n) =((3n)/2)(3+3)=9n
$${S}_{{n}} =\mathrm{2}{n} \\ $$$$\Rightarrow\frac{{n}}{\mathrm{2}}\left(\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right)=\mathrm{2}{n} \\ $$$$\Rightarrow\mathrm{2}{a}+{nd}−{d}−\mathrm{4}=\mathrm{0} \\ $$$${S}_{\mathrm{2}{n}} =\mathrm{5}{n} \\ $$$$\Rightarrow\frac{\mathrm{2}{n}}{\mathrm{2}}\left(\mathrm{2}{a}+\left(\mathrm{2}{n}−\mathrm{1}\right){d}\right)=\mathrm{5}{n} \\ $$$$\Rightarrow\mathrm{2}{a}+\mathrm{2}{nd}−{d}−\mathrm{5}=\mathrm{0} \\ $$$$ \\ $$$${subtracting}\:{the}\:{red}\:{ones} \\ $$$${nd}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{nd}=\mathrm{1} \\ $$$$ \\ $$$${putting}\:{value}\:{of}\:{nd}\:{in}\:{either}\:{of}\:{red}\:{lines} \\ $$$$\mathrm{2}{a}−{d}−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{a}−{d}=\mathrm{3} \\ $$$$ \\ $$$${S}_{\mathrm{3}{n}} =\frac{\mathrm{3}{n}}{\mathrm{2}}\left(\mathrm{2}{a}+\left(\mathrm{3}{n}−\mathrm{1}\right){d}\right) \\ $$$$=\frac{\mathrm{3}{n}}{\mathrm{2}}\left(\mathrm{2}{a}−{d}+\mathrm{3}{nd}\right) \\ $$$${using}\:{blue}\:{ones} \\ $$$${S}_{\mathrm{3}{n}} =\frac{\mathrm{3}{n}}{\mathrm{2}}\left(\mathrm{3}+\mathrm{3}\right)=\mathrm{9}{n} \\ $$
Commented by Kunal12588 last updated on 29/May/19
S_(pn) =((pn)/2)(2a+(pn−1)d)=((p(p+3)n)/2)  ; p∈N
$${S}_{{pn}} =\frac{{pn}}{\mathrm{2}}\left(\mathrm{2}{a}+\left({pn}−\mathrm{1}\right){d}\right)=\frac{{p}\left({p}+\mathrm{3}\right){n}}{\mathrm{2}}\:\:;\:{p}\in\mathbb{N} \\ $$
Commented by Tawa1 last updated on 29/May/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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