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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Evaluate `sum2^(i)`, where `i=2, 3, 4, … , 10`.A. 2044B. 2048C. 1024D. 1022 |
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Answer» Correct Answer - A `sum 2^(i), i =2, 3, ...10` `= (2^(2) + 2^(3)+... + 2^(10))` `= 2^(2) xx ((2^(9)-1))/(2-1)` `= (2^(11) -2^(2))/(1)` `= 2044`. |
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| 2. |
Evaluate `sum(3+2^(r ))`, where `r=1, 2, 3, …, 10`.A. 2051B. 2049C. 2076D. 1052 |
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Answer» Correct Answer - C `sum (3+2r) = 3 sum1 + overset(10) underset(r=1)(sum)2^(r )`. |
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| 3. |
Find the sum of the first 10 terms of geometric progression `18, 9, 4.5, …`A. `9((2^(10)-1))/(2^(8))`B. `9((2^(10)-1))/(2^(10))`C. `36((2^(10)-1)/(2^(8)))`D. `8((2^(10)-1))/(2^(8)) ` |
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Answer» Correct Answer - A Use the formula of `S_n` of a GP. |
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| 4. |
Find the sum of the series `1+ (1+2) + (1+2+3) + (1+2 + 3 +4)+ … + (1+2+3+… + 20)`.A. 1470B. 1540C. 1610D. 1370 |
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Answer» Correct Answer - B `1+(1+2)+ (1+2+3)+(1+2+3+4) +...20` terms. `= 1+3+6+10 + 15 + 21 + 28 + 36 +m 45+55 + 66 + 78 + 91 + 105+ 120 + 136 + 153 + 171 + 190 + 210` `= 1540` Alternate method : `1+(1+2) + (1+2 + 3) + ...n`th term of the series is `1+2 +3 + 4 + ...n = (n(n+1))/(2)` Sum of `n` terms of the series `= sum t_n` `= sum (n(n+1))/(2) rArr = (1)/(2) sum (n^(2) + n)` ` = (1)/(2) ( sum n^(2) + sumn)` `= (1)/(2)[(n(n+1) (2n+1))/(6) + (n(n+1))/(2)]` `= (n(n+1))/(4) ((2n+1)/(3) +1)` `rArr (n(n+1)(2n+4))/(12) = (n(n+1)(n+2))/(6)` If `n= 20` `sum t_(20) = (20(21)(22))/(6)` `rArr (10xx 21xx 22)/(3) = 1540`. |
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| 5. |
If the 3rd, 7th and 11th terms of a geometric progression are `p, q and r` respectively, then the relation among `p, q and r` is ____.A. `p^(2) = qr`B. `r^(2) = qp`C. `q^(2) = p^(2r2)`D. `q^(2)= pr` |
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Answer» Correct Answer - D `p = ar_1^(2) , q = ar_1^(6) and r = ar_1^(10)` |
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| 6. |
Geometic mean of `5, 10 and 20` is ______. |
| Answer» Correct Answer - 10 | |
| 7. |
Find the sum fo the first 22 terms of an AP whose first term is 4 and the common difference is `(4)/(3)`. |
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Answer» Given that, `a= 4 and d = (4)/(3)`. We have, `S_n = (n)/(2) [ 2a + (n-1)d]` `S_(22) = ((22)/(2)) [(2)(4)+ (22 -1)((4)/(3))] = (11)(8+ 28) = 369`. |
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| 8. |
Divide 124 into four parts in such a way that they are in AP and the product of the first and the 4 th part is 128 less than the product of the 2nd and the 3 rd parts. |
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Answer» Let the four parts be `(a-3d), (a-d) and (a+3d)`. The sum of these four parts is 124, i.e., `4a = 124 rArr a = 31` `(a-3d) (a+ 3d) = (a-d)(a+d)-128` `rArr a^(2) - 9d^(2) = a^(2) - a^(2) - 128` `rArr 8a^(2)= 128 rArr d = pm 4`. As `a = 31`, taking `d=4`, the four parts are 19, 27, 35 and 43. |
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| 9. |
If the sum of 16 terms of an AP is 1624 and the first term is 500 times the common difference, then find the common difference.A. 5B. `(1)/(2)`C. `(1)/(5)`D. `2` |
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Answer» Correct Answer - C `S_n = (n)/(2)[2a + (n-1)d]` `(16)/(2) [2xx 500d + (16 -1)d] = 1624` `8[1000d + 15d] = 1624` `1015d = (1624)/(8)` `1015d = 203` `d = (203)/(1015) rArr d = (1)/(5)` |
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| 10. |
Find the first term and the common difference of an AP, if the 3rd term is 6 and 17 th term is 34. |
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Answer» If a is the first term and d is the common difference, then we have `" "a + 2d = 6" " `(1) `" " a + 16 d = 34" "` (2) On substracting Eq. (1) from Eq. (2), we get `" " 14 d = 28 rArr d = 2` Substituting the value of d in Eq. (1), we get `a= 2` `therefore a = 2 and d = 2`. |
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| 11. |
If `a, b and c` are in geometric progression then, `a^(2), b^(2) and c^(2)` are in ____ progression. |
| Answer» Correct Answer - geometric | |
| 12. |
If every term of a series in geometric progression is multiplied by a real number, then the resulting series also will be in geometric progression. [True/False] |
| Answer» Correct Answer - True | |
| 13. |
For a series in geometric progression, the first term is a and the second term is 3a. The common ratio of the series is _______. |
| Answer» Correct Answer - 3 | |
| 14. |
Find the 14 th terms of an AP whose first whose first term is 3 and the common difference is 2. |
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Answer» The nth term of an AP is given by `t_n = a+ (n-1)d`, where a is the first term and d is the common difference. `therefore t_(14) = 3 + (14-1) 2 = 29`. |
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| 15. |
Find the sum of all the multiples of 6 between 200 and 1100.A. 96750B. 95760C. 97560D. 97650 |
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Answer» Correct Answer - D Form the series and find the value of n and use the formula `S_n= (n)/(2) (2a +(n-1)d)`. |
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| 16. |
Third term of the sequence whose nth term is `2n+5` is ____. |
| Answer» Correct Answer - 11 | |
| 17. |
The first term and the mth term of a geometric progression are `a and n` respectively and its nth term is `m`. Then its `(m+1-n)`th term is _____.A. `(ma)/(n)`B. `(na)/(m)`C. `mna`D. `(mn)/(a)` |
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Answer» Correct Answer - B (i) Use the formula to find `n`th term of a GP. (ii) `ar^(m-1) = n,ar^(n-1) =m`. (iii) `(m+1 -n)`th term =`ar^(m-n)` |
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| 18. |
If the `k`th term of a HP is `lamdap` and the `lamda`th term is `kp and k ne lamda`, then the `p` th term is ______.A. `k^(2)lamda`B. `k^(2)p`C. `p^(2)k`D. `lamda k` |
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Answer» Correct Answer - D `t_n = (1)/(a+(n-1)d)` in HP. |
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| 19. |
In the series, `T_n = 2n + 5`, find `S_4`. |
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Answer» `T_n = 2n +5` `T_1 = 2(1) + 5 = 7` `T_2 = 2(2) + 5 = 9` `T_3 = 2(3) + 5 = 11` `T_4 = 2(4) + 5 = 13` `S_4 = T_1 + T_2 + T_3 + T_4 = 7 + 9 + 11 + 13 = 40`. |
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| 20. |
Insert three harmonic between `(1)/(12) and (1)/(20)`. |
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Answer» After inserting the harmonic means, let the harmonic progression be, `(1)/(a) , (1)/(a+d) , (1)/(a+2d), (1)/(a+3d), (1)/(a+4d)` Given, `(1)/(a) = (1)/(12) and (1)/(a+4d) = (1)/(20)rArr a = 12 and d = 2` `therefore ` The required harmonic means are `(1)/(14), (1)/(16) and (1)/(18)`. |
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| 21. |
If `|x| lt 1`, then find the sum of the serires ` 2+ 4x + 6x^(2) + 8x^(3) + ……..`. |
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Answer» Let `S= 2+ 4x + 6x^(2) + 8x^(3)+ … " "` (1) `x S = 2x + 4x^(2)+ 6x^(3)+…" " `(2) Eq. (1) - Eq. (2) gives `S(1-x) = 2+ 2x + 2x^(3)+ 2x^(3)+ …` `" " = 2(1+x+x^(2) + …)` `1+ x + x^(2) +... ` is an infinite GP with `a = 1, r =x and |r| = |x| lt 1` `therefore ` Sum of the series = `(1)/(1-x)` `therefore S = (2)/((1-x)^(2))` . |
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| 22. |
If `T_n = 3n +8`, then `T_(n-1) =` ________. |
| Answer» Correct Answer - `3n+5` | |
| 23. |
The nth term of the sequence `(1)/(100), (1)/(10000), (1)/(1000000)`, ... is __________. |
| Answer» Correct Answer - `(1)/(100^(n))` | |
| 24. |
Find the 10 th term of the HP `(3)/(2), 1, (3)/(4), (3)/(5), ...` |
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Answer» The given HP is `(3)/(2), 1, (3)/(4), (3)/(5),...` The corresponding AP is `(2)/(3), 1, (4)/(3), (5)/(3),...` Here `a = (2)/(3) , d = 1 - (2)/(3) = (1)/(3)` `therefore T_(10)` of the corresponding AP is `a+ (10-1) d = (2)/(3) = (9)(1)/(3) = (11)/(3)` Hence, required term in HP is `(3)/(11)`. |
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| 25. |
The second and fifth term of GP are 3 and 81/8 respectively. Find the GP |
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Answer» Given T2 = ar = 3, T5 = ar4 = 81/8 T5/T2 = ar4/ar = r3 = (81/8)/3 = 27/8 ∴ r = 3√(33/23) = 3/2 If ar = 3 ⇒ a. 3/2 = 3 ⇒ a = 2 G.p is 2, 3, 9/2, ……. |
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| 26. |
Find the common ratio of the GP whose first and last terms are 25 and `(1)/(625)` respectively and the sum of the GP is `(19531)/(625)`. |
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Answer» We know that the sum of a GP is `("first term" - r("last term"))/(1-r)` or `(19531)/(625) = (25-((r )/(625)))/(1-r)` `therefore r= (1)/(5)` |
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| 27. |
Find the sum of the series `1, (2)/(5), (4)/(25), (8)/(125),… prop`. |
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Answer» Given that, `a =1, r = (2)/(5) and |r| = |(2)/(5)| lt 1` `therefore S_(oo) = (a)/(1-r) = (1)/(1-(2)/(5)) = (5)/(3)` |
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| 28. |
Ina series, `T_n = x^(2n-2) (x ne 0)`, then write the infinite series. |
| Answer» Correct Answer - `1+x^(2) + x^(4) + x^(6)+...` | |
| 29. |
The product of three numbers of a GP is `(64)/(27)`. If the sum of their products when taken in pairs is `(148)/(27)`, then find the numbers. |
| Answer» Correct Answer - `1, (4)/(3), (16)/(9)` | |
| 30. |
`S_(10)` is the sum of first 10 terms of a GP and `S_5` is the sum of the first 5 terms of the same GP. If `(S_(10))/(S_5)= 244`, then find the common ratio.A. 3B. 4C. 5D. 2 |
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Answer» Correct Answer - A `(S_(10))/(S_5) = (((a*(r^(10) -1)))/((r-1)))/(((a(r^(5)-1)))/(r-1))` or, `(S_(10))/(S_5) = (r^(10) -1)/(r^(5)-1) = r^(5) + 1` `" "(S_(10))/(S_5) = ((r^(5)-1)^(2) + 2r^(5)-2)/(r^(5)-1)` `244 = ((r^(5)-1)[r^(5)-1=2])/((r^(5)-1))rArr 244 = r^(5) +1` or, `r^(5) = 243 rArr r = 3`. |
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| 31. |
The harmonic mean of `1, 2 and 3` is `(3)/(2)`. [True/False] |
| Answer» Correct Answer - False | |
| 32. |
If the sum of first `n` terms which are in GP is `a(r+1)`, then the number of terms is _____. (Where a is the first term and r is the common ratio) |
| Answer» Correct Answer - 2 | |
| 33. |
In a GP of `7` terms, the last term is `(64)/(81)` and the common ratio is `(2)/(3)`. Find the 3rth term.A. 4B. 9C. 8D. 12 |
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Answer» Correct Answer - A Let `a` be the first term and r the common ratio of the GP. `T_y = ar^(6) = (64)/(81) rArr a =9` Therefore the 3rd term, `T_3 = ar^(2)` `T_3 = 9((2)/(3))^(2) =4`. |
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| 34. |
In a GP, if the fourth terms is the square of the second term, then the relation between the first term and common ratio is ____.A. `a =r`B. `a = 2r`C. `2a = r`D. `r^(2) = a` |
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Answer» Correct Answer - A `t_4 = t_2^(2)` `rArr ar^(3) = (ar)^(2) rArr ar^(3) = a^(2)r^(2) rArr r =a`. |
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| 35. |
If three arithemetic means are inserted between 4 and 5, then the commom difference is ______. |
| Answer» Correct Answer - `(1)/(4)` | |
| 36. |
Find the HM of `(1)/(7) and (1)/(12)`.A. `(1)/(19)`B. `(2)/(19)`C. `(3)/(19)`D. `(4)/(19)` |
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Answer» Correct Answer - B HM of `a and b` is `(2ab)/(a+b)` |
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| 37. |
If the 5th term and the 14 th term of an AP are 35 and 8 respectively, then find the 20 th term of the AP. |
| Answer» Correct Answer - `-10` | |
| 38. |
If the 7 th and the 9 th terms of a GP are `x and y` respectively, then the common ratio of the GP is ______. |
| Answer» Correct Answer - `pm sqrt((y)/(x))` | |
| 39. |
The AM of two numbers is 40 more than GM and 64 more than HM. Find the numbers. |
| Answer» Correct Answer - 180 and 20 | |
| 40. |
If the AM of two numbers is 9 and their HM is 4, then their GM is 6. [True/False] |
| Answer» Correct Answer - True | |
| 41. |
Find the sum to `n` terms of the series `5 + 55 + 555 + ...` |
| Answer» Correct Answer - `(50)/(81)(10^(n)-1)- (5n)/(9)` | |
| 42. |
The numbers `h_1, h_2, h_3, h_4,..., h_10` are in harmonicprogression and `a_1, a_2, ..., a_10` are in arithmetic progression. If `a_1 = h_1=3 and a_7=h_7=39`, thenthe value of `a_4 xx h_4` isA. `(13)/(49)`B. `(182)/(3)`C. `(7)/(13)`D. `117` |
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Answer» Correct Answer - D (i) In `AP, t_n = a + (n-1)d` and in HP, `t_n = (1)/(a+(n-1)d)`. (ii) Find the values of a and d by using the data. |
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| 43. |
If the sum of three consecutive terms of an AP is 9m, then the middle term is ____. |
| Answer» Correct Answer - 3 | |
| 44. |
If the ratio of the arithmatic mean and the geometric mean of two positive numbers is `3:2`, then find the ratio of the geometric mean and the harmonic mean of the numbers.A. `2:3`B. `9:4`C. `3:2`D. `4:9` |
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Answer» Correct Answer - C Use the relation between AM, GM and HM. |
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