Question 662463
The sum s of an arithmetic series with first term {{{a}}}, {{{n}}} terms, and common difference {{{d}}} is:


{{{s = n( 2a + (n - 1)d ) / 2}}}


Applying this to the given series:


{{{945 = n( 2 * 5 + 4(n - 1) ) / 2}}}

{{{945 = n( 10 + 4n - 4 ) / 2}}}


{{{945 = 2n( 4n +6 ) / 2}}}


{{{945 = 2n( 2n +3 ) / 2}}}

{{{945 = cross(2)n( 2n +3 ) /cross( 2)}}}


{{{945 = n( 2n +3 ) }}}


{{{945 = 2n^2 +3n  }}}


{{{2n^2 + 3n - 945 = 0}}}...factor...replace {{{3n}}} with {{{-42n+ 45n}}}


{{{2n^2 -42n+ 45n - 945 = 0}}}....group


{{{(2n^2 -42n)+ (45n - 945) = 0}}}


{{{2n(n -21)+ 45(n - 21) = 0}}}


{{{(2n + 45)(n - 21) = 0}}}


The positive root {{{n = 21}}} gives the {{{number}}} of {{{terms}}}.

so, your sequence is:

{{{5}}}, {{{9}}}, {{{13}}},{{{17}}},{{{21}}},{{{25}}},{{{29}}},{{{33}}},{{{37}}},{{{41}}},{{{45}}},{{{49}}},{{{53}}},{{{57}}},{{{61}}},{{{64}}},{{{69}}},{{{73}}},{{{77}}},{{{81}}},{{{85}}}

the sum is: 

{{{5+9+13+17+21+25+29+33+37+41+45+49+53+57+61+64+69+73+77+81+85 = 945}}}

{{{945 = 945}}}