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

The second, fourth, and eighth terms of an arithmetic progression (AP) with a non-zero common difference form three consecutive terms of a geometric progression (GP). If the sum of the first 55 terms of the AP is 4545, what is the first term of the AP?

  1. 3Answer
  2. B
    5
  3. C
    9
  4. D
    15

Answer

The first term of the AP is 3.
By expressing the 2nd, 4th, and 8th terms as a+da+d, a+3da+3d, and a+7da+7d, the geometric mean condition (a+3d)2=(a+d)(a+7d)(a+3d)^2 = (a+d)(a+7d) reduces to d=ad = a. Substituting d=ad = a into the sum formula S5=5(a+2d)=45S_5 = 5(a+2d) = 45 gives 15a=4515a = 45, yielding a first term of 3.

Step-by-Step Solution

1
Express the 2nd, 4th, and 8th terms of the AP in terms of first term aa and common difference dd
T2=a+dT_2 = a + d, T4=a+3dT_4 = a + 3d, and T8=a+7dT_8 = a + 7d
The nn-th term of an AP is given by Tn=a+(n−1)dT_n = a + (n-1)d.
2
Apply the condition for consecutive terms of a GP
(a+3d)2=(a+d)(a+7d)  ⟹  a2+6ad+9d2=a2+8ad+7d2  ⟹  2d2=2ad  ⟹  d=a(a + 3d)^2 = (a + d)(a + 7d) \implies a^2 + 6ad + 9d^2 = a^2 + 8ad + 7d^2 \implies 2d^2 = 2ad \implies d = a
If three terms x,y,zx, y, z form a GP, then y2=xzy^2 = xz. Since d≠0d \neq 0, dividing by 2d2d gives d=ad = a.
3
Use the sum of the first 5 terms of the AP to set up an equation for aa and dd
S5=52[2a+4d]=45  ⟹  5(a+2d)=45  ⟹  a+2d=9S_5 = \frac{5}{2}[2a + 4d] = 45 \implies 5(a + 2d) = 45 \implies a + 2d = 9
The sum of the first nn terms of an AP is Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d].
4
Substitute d=ad = a into the sum equation to solve for aa
a+2(a)=9  ⟹  3a=9  ⟹  a=3a + 2(a) = 9 \implies 3a = 9 \implies a = 3
Substituting d=ad = a simplifies the linear equation to solve directly for aa.

Key Concept

Combining Arithmetic and Geometric Progression properties to set up and solve simultaneous equations.
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Topics

Arithmetic and Geometric Progressions (AP and GP)

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