Question Number 175180 by Rasheed.Sindhi last updated on 22/Aug/22
$${a}_{{n}} \:{is}\:{an}\:{AP}\:{and}\:{S}_{{n}} \:{is}\:{sum}\:{of}\:{n}\:{terms} \\ $$$${of}\:{this}\:{AP}. \\ $$$${Given}\:{that}\:{S}_{\mathrm{11}} −{S}_{\mathrm{7}} =\mathrm{72},\:{determine} \\ $$$${a}_{\mathrm{6}} +{a}_{\mathrm{13}} . \\ $$
Answered by som(math1967) last updated on 22/Aug/22
![S_(11) −S_7 =72 ⇒((11)/2)(2a_1 +10d)−(7/2)(2a_1 +6d)=72 [a_1 =1st term d=c.d] ⇒4a_1 +55d−21d=72 ⇒4a_1 +34d=72 ⇒2a_1 +17d=36 ⇒(a_1 +5d)+(a_1 +12d)=36 ⇒a_6 +a_(13) =36](Q175182.png)
$${S}_{\mathrm{11}} −{S}_{\mathrm{7}} =\mathrm{72} \\ $$$$\Rightarrow\frac{\mathrm{11}}{\mathrm{2}}\left(\mathrm{2}{a}_{\mathrm{1}} +\mathrm{10}{d}\right)−\frac{\mathrm{7}}{\mathrm{2}}\left(\mathrm{2}{a}_{\mathrm{1}} +\mathrm{6}{d}\right)=\mathrm{72} \\ $$$$\left[{a}_{\mathrm{1}} =\mathrm{1}{st}\:{term}\:\:{d}={c}.{d}\right] \\ $$$$\Rightarrow\mathrm{4}{a}_{\mathrm{1}} +\mathrm{55}{d}−\mathrm{21}{d}=\mathrm{72} \\ $$$$\Rightarrow\mathrm{4}{a}_{\mathrm{1}} +\mathrm{34}{d}=\mathrm{72} \\ $$$$\Rightarrow\mathrm{2}{a}_{\mathrm{1}} +\mathrm{17}{d}=\mathrm{36} \\ $$$$\Rightarrow\left({a}_{\mathrm{1}} +\mathrm{5}{d}\right)+\left({a}_{\mathrm{1}} +\mathrm{12}{d}\right)=\mathrm{36} \\ $$$$\Rightarrow{a}_{\mathrm{6}} +{a}_{\mathrm{13}} =\mathrm{36} \\ $$
Commented by Rasheed.Sindhi last updated on 22/Aug/22
$$\mathcal{T}{hanks}\:\boldsymbol{{sir}}! \\ $$
Answered by Rasheed.Sindhi last updated on 22/Aug/22
![AnOther way S_(11) −S_7 =a_8 +a_9 +a_(10) +a_(11) [a_1 ^(×) +a_2 ^(×) +...+a_7 ^(×) +a_8 +a_9 +a_(10) +a_(11) ] =(a+7d)+(a+8d)+(a+9d)+(a+10d) =4a+34d=2(2a+17d) =2( (a+5d)+(a+12d) ) =2(a_6 +a_(13) )=72 a_6 +a_(13) =72/2=36](Q175186.png)
$${AnOther}\:{way} \\ $$$${S}_{\mathrm{11}} −{S}_{\mathrm{7}} ={a}_{\mathrm{8}} +{a}_{\mathrm{9}} +{a}_{\mathrm{10}} +{a}_{\mathrm{11}} \\ $$$$\left[\overset{×} {{a}_{\mathrm{1}} }+\overset{×} {{a}_{\mathrm{2}} }+...+\overset{×} {{a}_{\mathrm{7}} }+{a}_{\mathrm{8}} +{a}_{\mathrm{9}} +{a}_{\mathrm{10}} +{a}_{\mathrm{11}} \right] \\ $$$$=\left({a}+\mathrm{7}{d}\right)+\left({a}+\mathrm{8}{d}\right)+\left({a}+\mathrm{9}{d}\right)+\left({a}+\mathrm{10}{d}\right) \\ $$$$=\mathrm{4}{a}+\mathrm{34}{d}=\mathrm{2}\left(\mathrm{2}{a}+\mathrm{17}{d}\right) \\ $$$$=\mathrm{2}\left(\:\left({a}+\mathrm{5}{d}\right)+\left({a}+\mathrm{12}{d}\right)\:\right) \\ $$$$=\mathrm{2}\left({a}_{\mathrm{6}} +{a}_{\mathrm{13}} \right)=\mathrm{72} \\ $$$${a}_{\mathrm{6}} +{a}_{\mathrm{13}} =\mathrm{72}/\mathrm{2}=\mathrm{36} \\ $$
Answered by behi834171 last updated on 22/Aug/22
![S_m −S_n =(m/2)((m−1)d+2a_1 )−(n/2)((n−1)d+2a_1 )= =d[(m^2 /2)−(n^2 /2)−(m/2)+(n/2)]+(m−n)a_1 = =((m−n)/2)[(m+n−1)d+2a_1 ]=((m−n)/(m+n))S_(m+n) ⇒S_(m+n) =((m+n)/(m−n))(S_m −S_n ) ⇒S_(18) =((11+7)/(11−7))×72=364 a_6 +a_(13) =a_1 +5d+a_1 +12d=17d+2a_1 = =(9/9)[(18−1)d+2a_1 ]=(S_(18) /9)=36 . ■](Q175197.png)
$${S}_{{m}} −{S}_{{n}} =\frac{{m}}{\mathrm{2}}\left(\left({m}−\mathrm{1}\right){d}+\mathrm{2}{a}_{\mathrm{1}} \right)−\frac{{n}}{\mathrm{2}}\left(\left({n}−\mathrm{1}\right){d}+\mathrm{2}{a}_{\mathrm{1}} \right)= \\ $$$$={d}\left[\frac{{m}^{\mathrm{2}} }{\mathrm{2}}−\frac{{n}^{\mathrm{2}} }{\mathrm{2}}−\frac{{m}}{\mathrm{2}}+\frac{{n}}{\mathrm{2}}\right]+\left({m}−{n}\right){a}_{\mathrm{1}} = \\ $$$$=\frac{{m}−{n}}{\mathrm{2}}\left[\left({m}+{n}−\mathrm{1}\right){d}+\mathrm{2}{a}_{\mathrm{1}} \right]=\frac{{m}−{n}}{{m}+{n}}{S}_{{m}+{n}} \\ $$$$\Rightarrow{S}_{{m}+{n}} =\frac{{m}+{n}}{{m}−{n}}\left({S}_{{m}} −{S}_{{n}} \right) \\ $$$$\Rightarrow{S}_{\mathrm{18}} =\frac{\mathrm{11}+\mathrm{7}}{\mathrm{11}−\mathrm{7}}×\mathrm{72}=\mathrm{364} \\ $$$${a}_{\mathrm{6}} +{a}_{\mathrm{13}} ={a}_{\mathrm{1}} +\mathrm{5}{d}+{a}_{\mathrm{1}} +\mathrm{12}{d}=\mathrm{17}{d}+\mathrm{2}{a}_{\mathrm{1}} = \\ $$$$=\frac{\mathrm{9}}{\mathrm{9}}\left[\left(\mathrm{18}−\mathrm{1}\right){d}+\mathrm{2}{a}_{\mathrm{1}} \right]=\frac{{S}_{\mathrm{18}} }{\mathrm{9}}=\mathrm{36}\:\:\:.\:\blacksquare \\ $$
Commented by Rasheed.Sindhi last updated on 23/Aug/22
$${Wonderful}\:\boldsymbol{{sir}}!\:\mathbb{T}\mathrm{hank}\:\mathrm{you}. \\ $$