The sum of the first ten terms of an A.P. is | vedclass

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(A) Let the $A$.$P$. be $a, a+d, \dots$ and the $G$.$P$. be $g, gr, \dots$.Given that the sum of the first ten terms of the $A$.$P$. is $160$, we have

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$\frac{34}{9}$

Vedclass · Answer

(A) Let the $A$.$P$. be $a, a+d, \dots$ and the $G$.$P$. be $g, gr, \dots$.Given that the sum of the first ten terms of the $A$.$P$. is $160$, we have $S_{10} = \frac{10}{2}(2a + 9d) = 160$, which simplifies to $2a + 9d = 32$.Given that the sum of the first two terms of the $G$.$P$. is $8$, we have $g + gr = 8$, or $g(1+r) = 8$.We are given $a = r$ and $g = d$. Substituting these into the equations:$2r + 9g = 32$ and $g(1+r) = 8$.From the second equation, $g = \frac{8}{1+r}$. Substituting this into the first equation:$2r + 9(\frac{8}{1+r}) = 32 \Rightarrow 2r(1+r) + 72 = 32(1+r)$.$2r^2 + 2r + 72 = 32 + 32r \Rightarrow 2r^2 - 30r + 40 = 0 \Rightarrow r^2 - 15r + 20 = 0$.Let the roots be $r_1$ and $r_2$. Then $r_1 + r_2 = 15$ and $r_1r_2 = 20$.The first term of the $G$.$P$. is $g = \frac{8}{1+r}$. The sum of all possible values of $g$ is:$g_1 + g_2 = \frac{8}{1+r_1} + \frac{8}{1+r_2} = 8 \left( \frac{1+r_2 + 1+r_1}{(1+r_1)(1+r_2)} \right) = 8 \left( \frac{2 + (r_1+r_2)}{1 + (r_1+r_2) + r_1r_2} \right)$.Substituting the values: $8 \left( \frac{2 + 15}{1 + 15 + 20} \right) = 8 \left( \frac{17}{36} \right) = \frac{34}{9}$.

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