Question 845973
Let the first term of the AP be 'a' and the common difference be 'd'.
Therefore the n-th term is {{{t[n]=a+(n-1)d}}}
The sum up to first n-terms is {{{S[n]=n(2a+(n-1)d)/2}}}


Here, n=9, 


9th term is given by
{{{t[9]=42}}}
{{{a+(9-1)d = 42}}}
{{{a+8d=42}}} _______ (1)


Sum of first 9 terms is given by
{{{S[9]=414}}}
{{{9(2a+(9-1)d)/2=414}}}
{{{9(2a+8d)/2=414}}}
{{{9(a+4d)=414}}}
{{{a+4d=414/9}}}
{{{a+4d=46}}} _______ (2)


Subtracting both sides of equation (2) from (1)
{{{4d=-4}}}
{{{d=-1}}}


Substituting {{{d=-1}}} in (2) we have
{{{a-4=46}}}
{{{a=50}}}


First term of the sequence is 50.

Cheers!!!