Question 1186962
 
a) Find a in terms of d



Sum of {{{n }}}terms :

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

For {{{25}}} terms

{{{S[25]=(25/2)(2a+(25-1)d)}}}
{{{S[25] = 25a + 300d}}}


For {{{4}}} terms

{{{S[4]=(4/2)(2a+(4-1)d)}}}
{{{S[4] = 2(2a+3d)}}}
{{{S[4] = 4a + 6d}}}

given that the sum of the first {{{25}}} terms is {{{15 }}}times the sum of the first {{{4}}} terms, so e have

{{{25a + 300d = 15(4a + 6d)}}}
{{{25a + 300d = 60a + 90d}}}
 {{{300d- 90d = 60a -25a }}}
{{{35a =  210d}}}
{{{a =  210d/35}}}
{{{a =  6d}}}


b) Find the {{{55}}}th term in terms of {{{a}}}

{{{a =  6d}}}=> {{{d=  a/6}}}

{{{a[55]= a+(55-1)(a/6)}}}
{{{a[55]=a+54a/6}}}
{{{a[55]=a+9a}}}
{{{a[55]=10a}}}