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;

∴ S_n  = 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, …ar^{n – 1} and A, AR, AR², … AR^{n – 1} form a G.P, and find the common ratio.

Solution:

Product of sequence = a.A, ar.AR, ar².AR² ………….. ar^{n -1}.AR^{n -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 p^th, q^th and r^th terms of a G.P. are a, b and c, respectively. Prove that

a^{q – r} b^{r – p} c^{p – q} = 1.

Solution:

Let first term be A and common ratio be R.

T_p = AR^{p – 1}  = a -------------- (I)

T_q = AR^{q – 1} = b -------------- (II)

T_r = AR^{r – 1} = c --------------- (III)

a^{q – r} . b^{r – p} . c^{p – q} = (AR^{p – 1} )^{q – r} . (AR^{q – 1})^{r – p}  . (AR^{r – 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 n^th 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.