Question

Let \( a_1, a_2, a_3, \dots \) be an A.P. and \( g_1 = a_1, g_2 = a_2, g_3 = a_3, \dots \) be an increasing G.P. If \( a_1 = a_2 + g_2 = 1 \) and \( a_3 + g_3 = 4 \), then \( a_{10} + g_5 \) is equal to:

• A. 62
• B. 76
• C. 55
• D. 63.1

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
An Arithmetic Progression (A.P.) has a common difference \(d\), so \(a_n = a + (n-1)d\).
A Geometric Progression (G.P.) has a common ratio \(r\), so \(g_n = a \cdot r^{n-1}\).
We can use the given equations to find the first term \(a\), the difference \(d\), and the ratio \(r\).
Step 2: Key Formula or Approach:
Let the first term be \(a_1 = g_1 = a\).
From the problem, \(a_1 = 1\), so \(a = 1\).
The first given condition is \(a_2 + g_2 = 1\).
The second given condition is \(a_3 + g_3 = 4\).
Step 3: Detailed Explanation:
Substitute the general forms into the first condition.
\[ (a + d) + ar = 1 \] Since \(a = 1\), this simplifies nicely.
\[ (1 + d) + r = 1 \implies d + r = 0 \implies d = -r \] Now, substitute the general forms into the second condition.
\[ (a + 2d) + ar^2 = 4 \] Substitute \(a = 1\) and \(d = -r\).
\[ 1 + 2(-r) + r^2 = 4 \] \[ r^2 - 2r - 3 = 0 \] Factoring the quadratic equation yields the possible values for \(r\).
\[ (r - 3)(r + 1) = 0 \] This gives \(r = 3\) or \(r = -1\).
Since the G.P. is increasing and \(g_1 = 1>0\), the common ratio must be greater than 1.
Therefore, we select \(r = 3\).
This means \(d = -r = -3\).
We need to find the value of \(a_{10} + g_5\).
Calculate the 10th term of the A.P.
\[ a_{10} = a + 9d = 1 + 9(-3) = 1 - 27 = -26 \] Calculate the 5th term of the G.P.
\[ g_5 = a \cdot r^4 = 1 \cdot (3)^4 = 81 \] Finally, compute the required sum.
\[ a_{10} + g_5 = -26 + 81 = 55 \] Step 4: Final Answer:
The sum \(a_{10} + g_5\) is \(55\).