# CLASS-10GEOMETRIC PROGRESSION - PROBLEM & SOLUTION

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