Q13. How many terms of G.P. 3, 32, 33, …. are needed to give the sum 120?
Q14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Solution:
Q15. Given a G.P. with a = 729 and 7th term 64, determine S7.
Solution:
Q16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
Solution:
S2 = - 4,
T5 = 4(T3)
Q17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Solution:
T4 = ar3 = x ------------------ (I)
T10 = ar9 = y ---------------(II)
T16 = ar15 = z ---------------(III)
If T4, T10 and T16 are in G.P then x, y and z also will be in G.P
Q18. Find the sum to n terms of the sequence, 8, 88, 888, 8888… .
Solution:
Let S is the sum of n terms of series;
∴ Sn = 8 + 88 + 888 + 8888 + ………….. to the n term
= 8(1 + 11 + 111 + 1111 + ……….. )
Q20. Show that the products of the corresponding terms of the sequences a, ar, ar2, …arn – 1 and A, AR, AR2, … ARn – 1 form a G.P, and find the common ratio.
Solution:
Product of sequence = a.A, ar.AR, ar2.AR2 ………….. arn -1.ARn -1
Common ratio :
Q21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Solution:
Q22. If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that
aq – r br – p cp – q = 1.
Solution:
Let first term be A and common ratio be R.
Tp = ARp – 1 = a -------------- (I)
Tq = ARq – 1 = b -------------- (II)
Tr = ARr – 1 = c --------------- (III)
aq – r . br – p . cp – q = (ARp – 1 )q – r . (ARq – 1)r – p . (ARr – 1)p – q
= AR(p – 1)(q – r ).AR(q – 1)(r – p) . AR(r – 1)(p – q)
= AR(p – 1)(q – r ) + (q – 1)(r – p) + (r – 1)(p – q)
= AR(pq – pr – q + r + qr – pq – r + p + pr – qr – p + q)
= (AR)0
= 1
Q23. If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.
Solution:
Q24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is 1/rn.
Solution:
Q25. If a, b, c and d are in G.P. show that
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2.
Solution:
a, b, c, d are in G.P. Therefore,
bc = ad ……………………... (1)
b2 = ac …………………….... (2)
c2 = bd …………………...... (3)
It has to be proved that,
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2
R.H.S.
= (ab + bc + cd)2
= (ab + ad + cd)2 [Using (1)]
= [ab + d (a + c)]2
= (ab)2 + 2(ab)d(a + c) + [d(a + c)]2
= a2b2 + 2abd (a + c) + d2 (a + c)2
= a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2)
[Using (1) and (2)]
= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 [using bc = ad and b2 = ac]
= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2
= a2b2 + a2c2 + (ac)2 + b2c2 + b2c2 + a2d2 + (bd)2 × (bd)2 + c2d2
= a2b2 + a2c2 + (b2)2 + b2c2 + b2c2 + a2d2 + (bd)2 × (c2)2 + c2d2
= a2b2 + a2c2 + b2 × b2 + b2c2 + c2b2 + a2d2 + b2d2 + c2 × c2 + c2d2
= a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2
[Using (2) and (3) and rearranging terms]
= a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2)
= (a2 + b2 + c2) (b2 + c2 + d2) = L.H.S.
∴ L.H.S. = R.H.S.
∴ (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2
Q26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Solution:
Let G1, G2 be three numbers between 3 and 81 such that 3, G1, G2, 81 is a G.P
T1 = 3
T2 = ar
T3 = ar2
T4 = ar3 = 81
3.r3 = 81
r3 =
r3 = 27
For r = 3, we have
T2 = ar = 3.3 = 9
T3 = ar2 = 3.32 = 27
Q28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2√2) : (3 − 2√2).
Solution:
Let the numbers be a and b.