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a-n-is-an-AP-and-S-n-is-sum-of-n-terms-of-this-AP-Given-that-S-11-S-7-72-determine-a-6-a-13-




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_(11) −S_7 =72, determine  a_6 +a_(13) .
anisanAPandSnissumofntermsofthisAP.GiventhatS11S7=72,determinea6+a13.
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
S11S7=72112(2a1+10d)72(2a1+6d)=72[a1=1sttermd=c.d]4a1+55d21d=724a1+34d=722a1+17d=36(a1+5d)+(a1+12d)=36a6+a13=36
Commented by Rasheed.Sindhi last updated on 22/Aug/22
Thanks sir!
Thankssir!
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
AnOtherwayS11S7=a8+a9+a10+a11[a1×+a2×++a7×+a8+a9+a10+a11]=(a+7d)+(a+8d)+(a+9d)+(a+10d)=4a+34d=2(2a+17d)=2((a+5d)+(a+12d))=2(a6+a13)=72a6+a13=72/2=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   . ■
SmSn=m2((m1)d+2a1)n2((n1)d+2a1)==d[m22n22m2+n2]+(mn)a1==mn2[(m+n1)d+2a1]=mnm+nSm+nSm+n=m+nmn(SmSn)S18=11+7117×72=364a6+a13=a1+5d+a1+12d=17d+2a1==99[(181)d+2a1]=S189=36.◼
Commented by Rasheed.Sindhi last updated on 23/Aug/22
Wonderful sir! Thank you.
Wonderfulsir!Thankyou.

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