Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5. Let the sum of its first five terms be 98/25. Then the sum of the first 21 terms of an AP, whose first term is 10ar, nth term is an and the common difference is 10ar2, is equal to
- 21 a11
- 22 a11
- 15 a16
- 14 a16
The Correct Option is A
Solution and Explanation
The correct answer is (A) : 21 a11
Let first term of G.P. be a and common ratio is r
Then, \(\frac{α}{1-r} = 5....(i)\)
\(α =\frac{(r^5-1)}{(r-1)}\)
\(= \frac{98}{25}\)
\(⇒ 1-r^5 = \frac{98}{125}\)
\(∴ r^5 = \frac{27}{125}\)
\(r = (\frac{3}{5})^{\frac{3}{5}}\)
∴ Then, \(S_{21} = \frac{21}{2}[2×10ar + 20×10ar^2]\)
\(= 21[10ar+10.10ar^2]\)
\(= ^{21}α_{11}\)
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Concepts Used:
Geometric Progression
What is Geometric Sequence?
A geometric progression is the sequence, in which each term is varied by another by a common ratio. The next term of the sequence is produced when we multiply a constant to the previous term. It is represented by: a, ar1, ar2, ar3, ar4, and so on.
Properties of Geometric Progression (GP)
Important properties of GP are as follows:
- Three non-zero terms a, b, c are in GP if b2 = ac
- In a GP,
Three consecutive terms are as a/r, a, ar
Four consecutive terms are as a/r3, a/r, ar, ar3 - In a finite GP, the product of the terms equidistant from the beginning and the end term is the same that means, t1.tn = t2.tn-1 = t3.tn-2 = …..
- If each term of a GP is multiplied or divided by a non-zero constant, then the resulting sequence is also a GP with a common ratio
- The product and quotient of two GP’s is again a GP
- If each term of a GP is raised to power by the same non-zero quantity, the resultant sequence is also a GP.
If a1, a2, a3,… is a GP of positive terms then log a1, log a2, log a3,… is an AP (arithmetic progression) and vice versa