Note:
Anil Satpute
1) To find three consecutive terms in GP, consider the terms (a/r), (a), and (ar).
2) To find four consecutive terms in GP, consider the terms (a/r3), (a/r), (ar), and (ar3).
3) To find five consecutive terms in GP, consider the terms (a/r2), (a/r), (a), (ar), and (ar2).
A) Find three consecutive three terms in GP if the sum of the first and the third term is 35 and the product of all three terms is 2744.
Solution:
1) Let " a " be the first term and " r " be the common ratio of GP.
2) Let three consecutive terms in GP be (a/r), (a), and (ar).
3) According to the problem,
(a/r) (a)(ar) = 2744
(a3) = 4 x 686
(a3) = 4 x 2 x 343
(a3) = 4 x 2 x 7 x 49
(a3) = (2 x 7)3
(a) = (2 x 7)
(a) = (14)
4) Simillarly,
a/r + ar = 35
a (1/r + r) = 35
14 (1/r + r) = 35
(1/r + r) = 35/14
(1/r + r) = 5/2
(1/r + r) = 2 + 1/2
5) This shows that r = 2 or 1/2
6) Answer: Taking r = 2 and a = 14, the three consecutive terms are 7, 14, and 28.
Taking r = 1/2 and a = 14, the three consecutive terms are 28, 14, and 7.
Note:
1) Arithmetic Mean: The three terms p, q, and r in AP give " q " as the arithmetic means between p and q. So,
q - p = r - q
q + q = r + p
2 q = r + p
q = (r + p) / 2
2) Geometric Mean: The three terms p, q, and r in GP give " q " as the geometric mean between p and r. So,
q / p = r / q
q x q = r x p
q2 = r x p
q = (r p)1/2
Some Important Problems with AP & GP
A) In AP 6, 12, 18, ....., how many terms show the sum as 7650.
Solution:
1) Here, a = 6, d = 6 and Sn = 7650.
2) We know that
3) Answer: Total number of terms in AP having a sum of 7650 is 50.
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