PROBLEM & SOLUTION -
Example.1) Show the progression 81, 27, 9, 3, 1,…………..is a G.P. write its -
(i) First term (ii) Common ratio
(iii) e-th term (iv) 10th term
Ans.) The given progression is 81, 27, 9, 3, 1,…………..
27 9 3 1
We have, ------- = ------- = ------- = ------- (constant)
81 27 9 3
So, the given progression is a G.P.
We have –
(i) First term, a = 81 (Ans.)
27 1
(ii) common ratio, r = ------- = ------ …………….. (Ans.)
81 3
(iii) e-th term, Tₑ = arᵉˉ¹
1
=> Tₑ = 81 X (------)ᵉˉ¹
3
81
= -------- (Ans.)
3ᵉˉ¹
(iv) Putting e = 10 in Tₑ = arᵉˉ¹, where a = 81, r = 1/3, and we get –
Tₑ = arᵉˉ¹
1
=> Tₑ = 81 X (------)¹⁰ˉ¹
3
81
= -------
3⁹
3⁴
= --------
3⁹
1 1
= ------- = -------- (Ans.)
3⁵ 243
Example.2) If the 4th and 7th terms of a G.P. are 54 and 1458 respectively, then find its –
(i) common ratio, r (ii) first term, a
(iii) e-th term, Tₑ (iv) 5th term, T₅
Ans.) Let, a be the first term and r be the common ratio of the given G.P.
Then, T₄ = 54
=> T₄ = ar⁴ˉ¹ = ar³ = 54 ……………………(1)
And, T₇ = 1458
=> T₇ = ar⁷ˉ¹ = ar⁶ = 1458 …………………..(2)
On dividing the corresponding sides of (2) and (1), and we get –
ar⁶ 1458
-------- = ---------
ar³ 54
=> r⁶ˉ³ = 27
=> r³ = 3³
=> r = 3
So, the common ratio, r = 3 ………………………(i) (Ans.)
Now, we will put the value of r = 3, in (1), and we get –
ar³ = 54 ……………………(1)
=> a X 3³ = 54
=> a = 2 …………………………(ii) (Ans.)
Now, e-th term, Tₑ = arᵉˉ¹, we will substitute the value of a = 2, r = 3, and we get –
=> Tₑ = (2 X 3ᵉˉ¹) …………………….(iii) (Ans.)
5th term, T₅ = ar⁵ˉ¹, now we will substitute the value of a = 2, r = 3, and we get –
=> T₅ = [2 X 3⁽⁵ˉ¹⁾]
= (2 X 3⁴)
= (2 X 81)
= 162 ……………………..(iv) (Ans.)
1 1 1
Example.3) Find the sum of 7 terms of the G.P.-----, -----, -----,
3 6 12
1
------,…………….
24
1 1 1 1
Ans.) The given G.P. is ------, ------, ------, ------,…………….
3 6 12 24
1
In this given G.P., we have a = -------,
3
1/6 1 3 1
r = ------- = (------ X ------) = ------ < 1
1/3 6 1 2
And, e = 7,
a (1 - rᵉ)
So, using the formula, Sₑ = -------------, we get -
(1 – r)
1 1
------ X [1 – (------)⁷]
3 2
Sum of the 7 terms = S₇ = -----------------------------
1
(1 - -------)
2
1 2⁷ - 1
------ X (----------)
3 2⁷
= ---------------------------
2 - 1
(---------)
2
1 (128 – 1)
------ X -----------
3 128
= ------------------------------
1
-------
2
127 1 127 X 2 127
= -------- ÷ ------- = ------------ = --------
384 2 384 192
127
Hence the required sum is -------- (Ans.)
192
Your second block of text...