Cinemugam (Kollywood)

If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that x b−c × y c−a × z a−b = 1 . Solution : Since a, b and c are in A.P, b - a  = c - b  = d (common difference) We need to prove, x b−c × y c−a × z a−b = 1  Let us try to convert the powers in terms of one variable. 2b  = c + a - a + a 2b  = c - a + 2a 2(b - a)  = c - a 2d  = c - a  If c - b  = d, then b - c = -d If b - a  = d, then a - b = -d L.H.S x b−c × y c−a × z a−b  =  x −d × y 2d × z −d   ---(1) y = √xz By applying the value of y in (1)  =  x −d × (√xz) 2d × z −d    =  x −d × (xz) d × z −d     =  x −d + d   z -d + d   =  1 Hence proved.

Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years. Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years. What is his salary in the 4th year with respect to the offers A and B? Solution : Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years. a = 20000, r = 0.06 and n = 5 Starting salary   = 20,000  Every year 6% of annual salary is increasing. Starting salary  = 20000 Second year salary  = 20000 + 6% of 20000   =  20000(1 + 6%) By continuing in this way, we get 4th year salary  = 20000(1 + 6%) 3 =  20000(1.06) 3 =  22820 Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years. a = 22000, r = 0.03 and n = 5 Starting salary   = 22,000  Every year 3% of annual salary is increas

A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years? Solution : Starting salary   = 60,000  Every year 5% of annual salary is increasing. Second year salary  = 60000 + 5% of 60000   =  60000(1 + 5%) Third year salary  = 60000 + 5% of 60000(1 + 5%)   =  60000(1 + 5%)(1 + 5%)   =  60000(1 + 5%) 2 By continuing in this way,  salary after 5 years is   =  60000(1 + 5%)5   =  60000(105/100) 5   =  60000(1.05) 5    =  76577

In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 57/2 . Find the three terms. Solution : Let the three terms be a/r, a, ar The product of three consecutive terms  = 27 (a/r) ⋅ a ⋅ ar  = 27 a 3   = 27 a  = 3 Sum of the product of two terms taken at a time  = 57/2 [(a/r) ⋅ a] + [a ⋅ ar] + [ar ⋅ a/r]  = 57/2 a 2 /r + a 2 r + a 2   = 57/2 a 2 (1/r + r + 1)  = 57/2 9(1 + r + r 2 )/r  = 57/2 18(r 2 + r + 1)  = 57 r 18r 2 + 18r + 18  = 57 r 18r 2 + 18r - 57r + 18  = 0 18r 2  - 39r + 18  = 0 6r 2  - 13r + 6  = 0 (2r - 3)(3r - 2)  = 0 r = 3/2 and r = 2/3 First term  = a/r = 3/(3/2)  = 2 Second term  = a = 3 =  3 Third term  = ar = 3(3/2)  = 9/2 Hence the required three terms are 2, 3, 9/2 (or) 9/2, 3, 2.

If a, b, c are in A.P. then show that 3 a , 3 b , 3 c are in G.P. Solution : If a, b, c are in A.P, then b = (a + c)/2 In G.P,  b = √ac To prove that 3 a , 3 b , 3 c are in G.P. 3 b   = √( 3 a ⋅ 3 c )  3 b   = (3 a + c ) 1/2 3 b   = 3 (a+c)/2 b = (a + c)/2 Hence 3 a , 3 b , 3 c are in G.P.

Find the 10 th term of a G.P. whose 8 th term is 768 and the common ratio is 2. Solution : 8 th term is 768 t 8   = 768 ar 7   = 768 r = 2 a(2 7 )  = 768 a  = 768/128 a = 6 10th term : t 10   = ar 9 =  6(2 9 )   =  6(512) t 10  =  3072 Hence the 10 th term of the sequence is 3072.

In a G.P. the 9 th term is 32805 and 6 th term is 1215. Find the 12 th term. Solution : 9 th term is 32805 t 9   = 32805 ar 8   = 32805   -----(1) 6 th term is 1215 t 6   = 1215 ar 5   = 1215  -----(2) (2) / (1)  ==> ar 5 /ar 8   = 1215/32805 1/r 3   = 1/27 (1/r) 3   = (1/3) 3 r = 3 By applying the value of r in (2), we get a(3) 5   = 1215   a  = 1215/3 5 a  = 5(3 5 )/3 5 a = 5 12th term : t 12  =  ar 11 t 12   =  5(3) 11

Find the number of terms in the following G.P. (ii) 1/3, 1/9, 1/27,................1/2187 Solution : Let n th term be "1/2187" t n   = 1/2187 a = 1/3, r = (1/9)/(1/3)  = 1/3  ar n -1  =  1/2187 (1/3)(1/3) n -1  =  1/2187 (1/3) 1 + n-1  =  1/2187 (1/3) n   = (1/3) 7 n  = 7 Hence the 7th term of the above sequence is 1/2187. Hence there are 7 terms in the given G.P.

Find the number of terms in the following G.P. (i) 4, 8, 16,…,8192 ? Solution : Let n th term be "8192" t n   = 8192 a = 4, r = 8/4  = 2  ar n -1  =  8192 4(2) n -1  =  8192 2 2 (2) n -1  =  8192 2 n+1   = 2 13 n + 1  = 13 n = 12 Hence the 12 th term of the above geometric sequence is 8192. Hence there are 12 terms in the given G.P.

Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression. Solution : b  = √ac (x + 12)  = √(x + 6) (x + 15) Taking squares on both sides, (x + 12) 2   = (x + 6) (x + 15) x 2 + 12 2 + 2x(12)  = x 2 + 15x + 6x + 90 144 + 24x  = 21x + 90 24x - 21x  = 90 - 144 3x  = -54 x  = -18 Hence the value of x is -18.