Chapter 9 Sequences and Series Ex 9.3 Solutions
Question - 11 : - Evaluate: 
Answer - 11 : -
Question - 12 : - The sum of first three terms of a G.P.is 39/10 and their product is 1. Find the common ratio and the terms.
Answer - 12 : -
Let a/r, a, ar bethe first three terms of the G.P.
a/r + a + ar = 39/10…… (1)
(a/r) (a) (ar) = 1…….. (2)
From (2), we have
a3 = 1
Hence, a = 1[Considering real roots only]
Substituting the valueof a in (1), we get
1/r + 1 + r = 39/10
(1 + r + r2)/r= 39/10
10 + 10r + 10r2 =39r
10r2 –29r + 10 = 0
10r2 –25r – 4r + 10 = 0
5r(2r – 5) – 2(2r – 5)= 0
(5r – 2) (2r – 5) = 0
Thus,
r = 2/5 or 5/2
Therefore, the threeterms of the G.P. are 5/2, 1 and 2/5.
Question - 13 : - How many terms of G.P. 3, 32, 33, … are needed togive the sum 120?
Answer - 13 : -
Given G.P. is 3, 32,33, …
Let’s considerthat n terms of this G.P. be required to obtain the sum of120.
We know that,
Here, a =3 and r = 3
Equating the exponentswe get, n = 4
Therefore, four termsof the given G.P. are required to obtain the sum as 120.
Question - 14 : - The sum of first three terms of a G.P. is 16 and the sum of the nextthree terms is 128. Determine the first term, the common ratio and the sum to n termsof the G.P.
Answer - 14 : -
Let’s assume the G.P.to be a, ar, ar2, ar3,…
Then according to thequestion, we have
a + ar + ar2 =16 and ar3 + ar4 + ar5 =128
a (1 + r + r2)= 16 … (1) and,
ar3(1 + r + r2) = 128 … (2)
Dividing equation (2)by (1), we get
r3 = 8
r = 2
Now, using r = 2 in(1), we get
a (1 + 2 + 4) = 16
a (7) = 16
a = 16/7
Now, the sum of termsis given as
Question - 15 : - Given a G.P. with a = 729 and 7th term64, determine S7.
Answer - 15 : -
Given,
a = 729 and a7 =64
Let r bethe common ratio of the G.P.
Then we knowthat, an = a rn–1
a7 = ar7–1 = (729)r6
⇒ 64 = 729 r6
r6 = 64/729
r6 = (2/3)6
r = 2/3
And, we know that
Question - 16 : - Find a G.P. for which sum of the first twoterms is –4 and the fifth term is 4 times the third term.
Answer - 16 : -
Consider a tobe the first term and r to be the common ratio of the G.P.
Given, S2 =-4
Then, from thequestion we have
And,
a5 = 4x a3
ar4 =4ar2
r2 = 4
r = ± 2
Using the value of rin (1), we have
Therefore, therequired G.P is
-4/3, -8/3, -16/3, ….Or 4, -8, 16, -32, ……
Question - 17 : - If the 4th, 10th and 16th termsof a G.P. are x, y and z, respectively. Provethat x, y, z are in G.P.
Answer - 17 : -
Let a bethe first term and r be the common ratio of the G.P.
According to the givencondition,
a4 = a r3 = x …(1)
a10 = a r9 = y …(2)
a16 = ar15 = z … (3)
On dividing (2) by(1), we get
Question - 18 : - Find the sum to n terms of the sequence, 8, 88, 888,8888…
Answer - 18 : -
Given sequence: 8, 88,888, 8888…
This sequence is not aG.P.
But, it can be changedto G.P. by writing the terms as
Sn = 8 + 88 + 888 + 8888 + …………….. to n terms
Question - 19 : - Find the sum of the products of the corresponding terms of the sequences2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.
Answer - 19 : -
The required sum= 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x ½
= 64[4 + 2 + 1 + ½ +1/22]
Now, it’s seen that
4, 2, 1, ½, 1/22 isa G.P.
With first term, a =4
Common ratio, r =1/2
We know,
Therefore, therequired sum = 64(31/4) = (16)(31) = 496
Question - 20 : - Show that the products of the corresponding terms of thesequences a, ar, ar2, …arn-1 and A, AR, AR2,… ARn-1 form a G.P, and find the common ratio.
Answer - 20 : -
To be proved: Thesequence, aA, arAR, ar2AR2,…arn–1ARn–1, forms aG.P.
Now, we have
Therefore, the abovesequence forms a G.P. and the common ratio is rR.