Question

\( a_1, a_2, a_3, \dots, a_n \) are in A.P. and sum of first 10 terms is 160. \( g_1, g_2, g_3, \dots, g_n \) are in G.P. where \( g_1 + g_2 = 8 \). If the first term of A.P. is equal to common ratio of G.P. and first term of G.P. is equal to common difference of A.P., then sum of all possible values of \( g_1 \) is equal to:

• A. \( \frac{34}{9} \)
• B. \( \frac{28}{9} \)
• C. \( \frac{23}{3} \)
• D. \( \frac{28}{5} \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Let A.P. have first term \( a \) and difference \( d \). Let G.P. have first term \( G \) and ratio \( r \). Given: \( a = r \) and \( G = d \).
Step 2: Key Formula or Approach:
1. \( S_{10} = \frac{10}{2}[2a + 9d] = 160 \implies 2r + 9G = 32 \) --- (i)
2. \( g_1 + g_2 = 8 \implies G + Gr = 8 \implies G(1 + r) = 8 \) --- (ii)
Step 3: Detailed Explanation:
1. From (i), \( r = \frac{32 - 9G}{2} \). Substitute this into (ii): \[ G\left(1 + \frac{32 - 9G}{2}\right) = 8 \implies G\left(\frac{2 + 32 - 9G}{2}\right) = 8 \] \[ 34G - 9G^2 = 16 \implies 9G^2 - 34G + 16 = 0 \] 2. This is a quadratic in \( G \) (where \( G = g_1 \)). The sum of all possible values of \( g_1 \) is the sum of the roots of this equation. 3. Sum of roots \( = -b/a = \frac{34}{9} \).
Step 4: Final Answer:
The sum of all possible values of \( g_1 \) is \( \frac{34}{9} \).