The question is: the first three terms of an arithmetic series $c_{n}$ are $$a(1+b), a(1+3b),a(1+5b)$$ I needed to find the common difference in terms of $a$ and $b$ and then find the expression for $c_{n}$.
The final part I struggled with where I have to find $a$ and $b$ and the information given is $$c_{5} = 25,c_{10} = 55$$
The answers for the first two parts are $difference = 2ab$ and $c_{n}=a(1+(2n-1)b)$
On
On simplifying , the first three terms of series become a+ab , a+3ab , a+5ab .Therefore the common difference of the series is 2ab.
Now the nth term of the series is: (a+ab)+(n-1)(2ab).
on simplifying it becomes
a(1+(2n-1)b)
since the 5th and 10th term of series are 25 and 55.we get the relations:
25=a(1+9b) ... (1)
55=a(1+19b) ... (2)
dividing equation (1) by equation (2) , we get the value of b=-3/2 and substituting it in equation 1 we get the value of a = -2.
If there is any doubt,feel free to ask in comments.
The difference between the fifth term and the tenth term is five times the common difference, or $10ab$. Hence $10ab = 55-25 = 30$, so $ab=3$.
The fifth term is also $a(1+9b) = a+9ab = a+27 = 25$, so $a = -2$, which means $b = 3/a = -3/2$.
Thus $(a,b) = (-2, -3/2)$.