Mathematics - Sequence & Series Question with Solution | TestHub
Match the following and select the correct answer.
List I | List II |
|---|---|
P: Number of common integers in the A.P.s 1, 3, 5, 7, ..., 157 and 2, 5, 8, 11, ..., 161 is. | 1: 25 |
Q: If a, b are roots of 11x^2 - 4x - 2 = 0, then 10(1 + a + a^2 + ...)(1 + b + b^2 + ...) is, with both sums to infinity. | 2: 26 |
R: If the sum of the first 20 terms equals the sum of the first 10 terms in an A.P., then the sum of the first 30 terms is less than. | 3: 22 |
S: If 3/(1^2·2^2) + 5/(2^2·3^2) + ... + 199/(99^2·100^2) = a/b in lowest terms, then 20|b - a| is. | 4: 20 |
| 5: 10 |
Options:
Answer:
Solution:
Item , so and the common terms are with . The last term gives to 26 , so the count is matches 2 . Item Q: . Q matches 3. Item R: implies , which is less than each of 25, 26, 22, 20, 10. R matches 1,2,3,4,5. Item S: the sum telescopes to , so and . S matches 4 .
