Note:

Anil Satpute
1) To find three consecutive terms in GP, consider the terms as (a/r), (a), (ar).

2) To find four consecutive terms in GP, consider the terms as (a/r

^{3}), (a/r), (ar), (ar^{3}).
3) To find five consecutive terms in GP, consider the terms as (a/r

^{2}), (a/r), (a), (ar), (ar^{2}).
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 the 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), (ar).

3) According to the problem,

(a/r) (a)(ar) = 2744

(a

^{3}) = 4 x 686
(a

^{3}) = 4 x 2 x 343
(a

^{3}) = 4 x 2 x 7 x 49
(a

^{3}) = (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 gives " q " as the arithmetic mean 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 gives " q " as the geometric mean between p and r. So,

q / p = r / q

q x q = r x p

q

^{2}= r x p
q = (r p)

^{1/2}^{Some Important Problems on AP & GP}

^{ }

^{A) In an AP 6, 12, 18, ....., how many terms shows the sum as 7650.}

^{ }

^{Solution:}

1) Here, a = 6, d = 6 and S

_{n}= 7650.
2) We know that

3) Answer: Total number of term in AP having sum as 7650 is 50.