Question
The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8 . If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:
Options
Solution
Sol. [Key Concept: $S_{10} = 160$ gives $2a + 9d = 32$; the G.P. is $d,da,{da}^{2}$ with $d + da = 8$, so substitute to form a quadratic in $d$ and add its roots.]
Sum of 10 terms of A.P, $S_{10} = 160$
$${5\lbrack 2a + 9\text{ }d\rbrack = 160 }{2a + 9\text{ }d = 32 }$$G.P. $\Rightarrow d,da,{da}^{2}\ldots$.
$${\Rightarrow d + da = 8 }{\Rightarrow \frac{d + d(32 - 9\text{ }d)}{2} = 8 }{\Rightarrow 2\text{ }d + 32\text{ }d - 9{\text{ }d}^{2} = 8 }{\Rightarrow 9{\text{ }d}^{2} - 34\text{ }d + 8 = 0 }{d_{1} + d_{2} = \frac{34}{9} }$$
Create a free account to view solution
View Solution Free