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Difficulty: Very hardArithmetic and Geometric Progressions (AP and GP)

The sum of the first three terms of an arithmetic progression (AP) with a positive common difference dd is 2121. If 22 is added to the first term, 33 is added to the second term, and 99 is added to the third term, the resulting three numbers form consecutive terms of a geometric progression (GP). What is the sum of the first 1010 terms of this arithmetic progression?

Answer: 210

Answer

The sum of the first 10 terms of the arithmetic progression is 210.
Representing the AP terms as 7−d,7,7+d7-d, 7, 7+d and adding the specified values produces GP terms 9−d,10,16+d9-d, 10, 16+d. Solving 102=(9−d)(16+d)10^2 = (9-d)(16+d) yields d=4d=4. Consequently, the first term of the AP is 33. Using S10=102[2(3)+9(4)]S_{10} = \frac{10}{2}[2(3) + 9(4)] gives the final answer 210.

Step-by-Step Solution

1
Express AP terms symmetrically and solve for the middle term.
The middle term is a=7a = 7, making the terms 7−d7-d, 77, and 7+d7+d.
Choosing terms a−d,a,a+da-d, a, a+d allows the sum equation 3a=213a = 21 to directly isolate the middle term.
2
Set up the geometric progression relation to determine common difference dd.
The GP terms are 9−d9-d, 1010, and 16+d16+d. Solving 102=(9−d)(16+d)10^2 = (9-d)(16+d) gives d2+7d−44=0d^2 + 7d - 44 = 0, yielding d=4d = 4.
In any geometric progression, the square of the middle term equals the product of the first and third terms.
3
Determine the first term a1a_1 and calculate S10S_{10}.
The first term is a1=7−4=3a_1 = 7 - 4 = 3, and the sum S10=102[2(3)+(10−1)(4)]=210S_{10} = \frac{10}{2}[2(3) + (10-1)(4)] = 210.
Applying the AP sum formula Sn=n2[2a1+(n−1)d]S_n = \frac{n}{2}[2a_1 + (n-1)d] with n=10n=10, a1=3a_1=3, and d=4d=4.

Key Concept

Integrating AP and GP structural relationships to solve for sequence parameters and evaluate finite sums
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Topics

Arithmetic and Geometric Progressions (AP and GP)

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