Q5. Find the 20th and nth terms of the G.P. 5/2, 5/4, 5/8, … (NCERT Exercise 8.2, Q1).
a = 5/2, r = 1/2: a20 = (5/2)(1/2)19 = 5/220 and an = (5/2)(1/2)n−1 = 5/2n.
Q6. The 4th term of a G.P. is the square of its second term, and the first term is −3. Determine its 7th term (NCERT Exercise 8.2, Q4).
−3r3 = (−3r)2 = 9r2, so r = −3 (r ≠ 0). a7 = −3 × (−3)6 = −2187.
Q7. For what values of x are −2/7, x, −7/2 in G.P. (NCERT Exercise 8.2, Q6)?
x2 = (−2/7)(−7/2) = 1, so x = 1 or x = −1.
Q8. Find the sum of 20 terms of 0.15, 0.015, 0.0015, … (NCERT Exercise 8.2, Q7).
a = 0.15, r = 0.1: S20 = 0.15(1 − 0.120)/0.9 = (1/6)(1 − 0.120).
Q9. Find the sum to n terms of (i) 1, −a, a2, −a3, … (a ≠ −1); (ii) x3, x5, x7, … (x ≠ ±1) (NCERT Exercise 8.2, Q9 and Q10).
(i) r = −a: Sn = (1 − (−a)n)/(1 + a). (ii) r = x2: Sn = x3(1 − x2n)/(1 − x2).
Q10. Evaluate the sum of (2 + 3k) for k = 1 to 11 (NCERT Exercise 8.2, Q11).
The 2s add up to 22, and 3 + 32 + … + 311 = 3(311 − 1)/2 = 265 719. Total 265 741.
Q11. The sum of the first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms (NCERT Exercise 8.2, Q12).
Terms a/r, a, ar: a3 = 1, so a = 1, and 1/r + 1 + r = 39/10 gives 10r2 − 29r + 10 = 0, r = 5/2 or 2/5. The terms are 2/5, 1, 5/2 (or 5/2, 1, 2/5).
Q12. The sum of the first three terms of a G.P. is 16 and the sum of the next three terms is 128. Find the first term, the common ratio and the sum to n terms (NCERT Exercise 8.2, Q14).
The next three terms are the first three times r3, so r3 = 8 and r = 2. Then a(1 + 2 + 4) = 16 gives a = 16/7, and Sn = (16/7)(2n − 1).
Q13. Find a G.P. for which the sum of the first two terms is −4 and the fifth term is 4 times the third term (NCERT Exercise 8.2, Q16).
ar4 = 4ar2 gives r = 2 or −2. With r = 2: 3a = −4, a = −4/3, the G.P. −4/3, −8/3, −16/3, …. With r = −2: −a = −4, a = 4, the G.P. 4, −8, 16, ….
Q14. Find the sum to n terms of 8, 88, 888, 8888, … (NCERT Exercise 8.2, Q18).
As in Example 5:
Sn = (8/9)[10(10n − 1)/9 − n]
= (80/81)(10n − 1) − 8n/9.
Q15. Find the sum of the products of the corresponding terms of 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2 (NCERT Exercise 8.2, Q19).
The products 256, 128, 64, 32, 16 form a G.P. with r = 1/2: 256 + 128 + 64 + 32 + 16 = 496.
Q16. If the 4th, 10th and 16th terms of a G.P. are x, y and z, prove that x, y, z are in G.P. (NCERT Exercise 8.2, Q17).
x = ar3, y = ar9, z = ar15, so y/x = z/y = r6, and y2 = xz.
Q17. Find four numbers forming a G.P. in which the third term is greater than the first by 9 and the second term is greater than the fourth by 18 (NCERT Exercise 8.2, Q21).
a(r2 − 1) = 9 and ar(1 − r2) = 18. Dividing, −r = 2, so r = −2 and a = 3: 3, −6, 12, −24.
Q18. Insert two numbers between 3 and 81 so that the resulting sequence is a G.P. (NCERT Exercise 8.2, Q26).
81 = 3r3, so r = 3: the numbers are 9 and 27.
Q19. The number of bacteria in a culture doubles every hour. If there were 30 at first, how many are there at the end of the 2nd hour, the 4th hour and the nth hour (NCERT Exercise 8.2, Q30)?
30 × 22 = 120, 30 × 24 = 480 and 30 × 2n.
Q20. What will ₹500 amount to in 10 years in a bank that pays 10% interest a year, compounded annually (NCERT Exercise 8.2, Q31)?
Each year multiplies the amount by 1.1: 500 × 1.110 ≈ ₹1296.87.
Q21. The sum of some terms of a G.P. is 315, the first term is 5 and the common ratio is 2. Find the last term and the number of terms (NCERT Miscellaneous Exercise, Q2).
5(2n − 1) = 315 gives 2n = 64, n = 6, and the last term is 5 × 25 = 160.
Q22. A manufacturer reckons that a machine costing ₹15 625 will depreciate by 20% each year. Find its estimated value after 5 years (NCERT Miscellaneous Exercise, Q17).
Each year multiplies the value by 0.8: 15 625 × 0.85 = ₹5120.
Q23. If f satisfies f(x + y) = f(x)f(y) for all natural numbers x, y, f(1) = 3, and the sum of f(x) for x = 1 to n is 120, find n (NCERT Miscellaneous Exercise, Q1).
f(2) = f(1)2 = 9, and in general f(x) = 3x, a G.P. with a = r = 3. So 3(3n − 1)/2 = 120, 3n = 81 and n = 4.
Q24. The first term of a G.P. is 1, and the sum of its third and fifth terms is 90. Find the common ratio (NCERT Miscellaneous Exercise, Q3).
r2 + r4 = 90 gives (r2 + 10)(r2 − 9) = 0, so r2 = 9 and r = 3 or −3.
Q25. A G.P. has an even number of terms. The sum of all its terms is 5 times the sum of the terms in the odd places. Find the common ratio (NCERT Miscellaneous Exercise, Q5).
Each term in an even place is r times the term before it, so the even-place sum is r times the odd-place sum. For a real G.P. with non-zero first term, the odd-place sum is a(1 + r2 + …), which is non-zero. The total is (1 + r) times that sum, so cancellation gives 1 + r = 5 and r = 4.
Q26. Find the sum to n terms of (i) 5 + 55 + 555 + …; (ii) .6 + .66 + .666 + … (NCERT Miscellaneous Exercise, Q11).
(i) As in Example 5: (5/9)[10(10n − 1)/9 − n] = (50/81)(10n − 1) − 5n/9.
(ii) (6/9)[(1 − 0.1) + (1 − 0.01) + …] = (2/3)[n − (1/9)(1 − 10−n)].
Q27. A person writes a letter to four friends and asks each to copy it and mail it to four different people, who are to do the same. If the chain is not broken and each letter costs 50 paise to post, find the total amount spent on postage through the 8th set of letters (adapted from NCERT Miscellaneous Exercise, Q15).
The sets have 4, 16, 64, …, 48 letters: 4(48 − 1)/3 = 87 380 letters in all, costing 87 380 × ₹0.50 = ₹43 690.