Question 768612
 
26,21,16,11..........are needed to give the sum 11
26-21=5
21-16=5
16-11=5
so, d=-5

{{{Sn = n/2(2a + ( n-1 )d )}}} , where ‘n’ is the number of terms, ‘a’ is first term, ‘d’ is the common difference 

{{{a=26}}},{{{ Sn=11}}},{{{ d=-5}}}

{{{11 = n/2(2*26 + ( n -1 )(-5) )}}} ....solve for {{{n}}}

{{{11 = n/2(52 - 5n +5  )}}}

{{{11*2 = n(57 -5n  )}}}

{{{22 = 57n -5n^2  }}}

{{{ 5n^2-57n +22=0}}}....solve for {{{n}}} using quadratic formula

{{{n = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{n = (-(-57) +- sqrt( (-57)^2-4*5*22 ))/(2*5) }}}

{{{n = (57 +- sqrt( 3249-440 ))/10 }}}

{{{n = (57 +- sqrt( 2809 ))/10 }}}

{{{n = (57 +- 53)/10 }}}

{{{ n = 11}}} and {{{n = 2/5}}}


so, we need only whole number {{{ n = 11}}}


26,21,16,11,6,1,-4,-9,-14,-19,-24

{{{26+21+16+11+6+1+(-4)+(-9)+(-14)+(-19)+(-24)=11}}}