Question 1011604
this  answer  {{{a = 0}}}, {{{d = 12}}} is incorrect

to find nth term in an arithmetic sequence you use this formula:

 {{{a[n] = a + (n - 1)d}}}

and to find the sum of {{{n}}} terms in an arithmetic sequence you use this formula:

{{{S[n]=(n/2)(2a+(n-1)d)}}}


you are given {{{S[11] = 5580}}} (sum of {{{11}}} terms),and  {{{a[11] = 360}}} ({{{11th}}} term

 {{{360 = a + (11 - 1)d}}}

{{{360 = a + 10d}}}

{{{a=360-10d}}}........eq.1


{{{S[n]=(n/2)(2a+(n-1)d)}}}

{{{5580=(11/2)(2a+(11-1)d)}}}

{{{5580=(11/2)(2a+10d)}}}

{{{5580=(11/cross(2)1)(cross(2)a+cross(10)5d)}}}

{{{5580=11(a+5d)}}}

{{{5580/11=a+5d}}}

{{{a=5580/11-5d}}}.....eq.2


from eq.1 and eq.2 we have

{{{5580/11-5d=360-10d}}}.........solve for {{{d}}}

{{{10d-5d=360-5580/11}}}

{{{5d=(360*11)/11-5580/11}}}

{{{5d=3960/11-5580/11}}}

{{{5d=-1620/11}}}

{{{d=-1620/(5*11)}}}

{{{d=-cross(1620)324/(cross(5)*11)}}}

{{{highlight(d=-324/11)}}}

then {{{a}}} will be

{{{a=360-10(-324/11)}}}........eq.1

{{{a=360+3240/11}}}

{{{a=3960/11+3240/11}}}

{{{highlight(a=7200/11)}}}



check {{{11th}}} term:

{{{a[n] = a + (n - 1)d}}} if {{{a[11]=360}}},{{{a=7200/11}}} and {{{d=-324/11}}}

{{{360 = 7200/11 + (11- 1)(-324/11)}}}

{{{360 = 7200/11 + 10(-324/11)}}}

{{{360 = 7200/11-3240/11}}}

{{{360 = 3960/11}}}

{{{360 = 360}}} which confirms our solution

and

{{{S[n]=(n/2)(2a+(n-1)d)}}} if {{{S[11]=5580}}},{{{a=7200/11}}} and {{{d=-324/11}}}

{{{5580=(11/2)(2(7200/11)+(11-1)(-324/11))}}}

{{{5580=(11/2)(14400/11+10(-324/11))}}}

{{{5580=(11/2)(14400/11-3240/11)}}}

{{{5580=(11/2)(11160/11)}}}

{{{5580=(cross(11)1/2)(11160/cross(11)1)}}}

{{{5580=(1/2)11160}}}

{{{5580=11160/2}}}

{{{5580=5580}}} which confirms our solution