Respuesta :
Given that,
Series 1: 3, 5.5, 8..... and series 2: 33/2, 57/2, 81/2.
To find,
The value of n.
Solution,
Let n is the no of terms in both the series.
For 3, 5.5, 8
First term, a = 3 and common difference d = 2.5
Sum of n terms of a series is given by :
[tex]S=\dfrac{n}{2}(2a+(n-1)d)[/tex]
So,
[tex]S=\dfrac{n}{2}(2\times 3+(n-1)2.5)\\\\S=\dfrac{n}{2}(6+2.5(n-1))\ ....(1)[/tex]
For 33/2, 57/2, 81/2
First term, a = 33/2 and common difference, d = 12
nth term of a series is given by :
[tex]a_n=a+(n-1)d[/tex]
For 2nd terms,
[tex]a_2=\dfrac{33}{2}+(2-1)\times 12\ ....(2)[/tex]
According to question, sum of the first n terms of the arithmetical progression 3, 5.5, 8. .... is equal to the 2nth term of the arithmetic progression 33/2, 57/2, 81/2
From equation (1) and (2) :
[tex]\dfrac{n}{2}(6+2.5(n-1))=\dfrac{33}{2}+ 12\\\\\dfrac{n}{2}(6+2.5(n-1))=\dfrac{57}{2}\\\\n(6+2.5(n-1))=57\\\\6n+2.5n^2-2.5n-57=0\\\\2.5n^2+2.5n-57=0\\\\\text{On solving the above quadratic equation, we get the value of x as follows :}\\\\x=4.126,-5.526[/tex]
Neglecting the negative value, we get the value of n = 4.126
or
n = 4
Hence, the value of n is 4.
