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BBA-DBE Bangalore 2026 PYQsQuantitative Ability. Free, no login required.

The 5th term of an arithmetic sequence is 20 and its 10th term is 35. The first term of a geometric sequence is the same as the first term of the arithmetic sequence mentioned above, while the 4th term of the geometric sequence is 1. What will be the sum to infinity of the arithmetic-geometric sequence whose nn-th term is given by [a+(n−1)d]×rn−1[a+(n-1)d]\times r^{n-1}, where aa is the first term of both the arithmetic and the geometric sequences, dd is the common difference of the arithmetic sequence, and rr is the common ratio of the geometric sequence?

Solution

✅ Correct Option: 4

For the arithmetic sequence:

a+4d=20a+4d=20

a+9d=35a+9d=35

Subtracting the equations gives 5d=155d=15, so d=3d=3 and a=8a=8.

The geometric sequence has first term 8 and fourth term 1:

8r3=18r^3=1

r=12r=\dfrac{1}{2}

The required infinite sum is ∑n=1∞[a+(n−1)d]rn−1\sum_{n=1}^{\infty}[a+(n-1)d]r^{n-1}.

Using ∑n=1∞[a+(n−1)d]rn−1=a1−r+dr(1−r)2\sum_{n=1}^{\infty}[a+(n-1)d]r^{n-1}=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}:

81/2+3(1/2)(1/2)2=16+6=22\dfrac{8}{1/2}+\dfrac{3(1/2)}{(1/2)^2}=16+6=22.

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