Answer:
[tex]2[/tex]
Step-by-step explanation:
[tex]\text{Solution:}\\\text{Let the first term of the A.P. be }a\text{ and the common difference be }d.[/tex]
[tex]\text{The sum of first }n\text{ terms, }S_n=\dfrac{n}{2}[2a+(n-1)d][/tex]
[tex]\text{So, sum of first 6 terms, }S_6=\dfrac{6}{2}[2a+(6-1)d]\\\\\text{or, }42=3(2a+5d)\\\text{or, }6a+15d=42\\\text{or, }a+2.5d=7\\\text{or, }a=7-2.5d.........(1)[/tex]
[tex]\text{Also, the nth term of an AP is given by: }\\t_n=a+(n-1)d\\\text{So, 11th term of A.P., }t_{11}=a+10d\\\text{33th term of A.P., }t_{33}=a+32d[/tex]
[tex]\text{Given,}[/tex]
[tex]\dfrac{t_{11}}{t_{33}}=\dfrac{1}{3}\\\\\text{or, }\dfrac{a+10d}{a+32d}=\dfrac{1}{3}\\\\\text{or, }3a+30d=a+32d\\\\\text{or, }2a=2d\\\\\text{or, }a=d[/tex]
[tex]\text{From equation(1),}\\7-2.5d=d\\\text{or, }3.5d=7\\\text{or, }d=2=a\\\therefore\ \text{First term of the AP = 2}[/tex]
