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Each of the series 13 + 15 + 17+…. and 14 + 17 + 20+… is continued to 100 terms. Find howmany terms are identical between the two series.
- a)35
- b)34
- c)32
- d)33
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Each of the series 13 + 15 + 17+…. and 14 + 17 + 20+… is c...
The two series till their hundredth terms are 13, 15, 17….211 and 14, 17, 20…311. The common
terms of the series would be given by the series 17, 23, 29….209. The number of terms in this
series of common terms would be 192/6 + 1 = 33. Option (d) is correct.
terms of the series would be given by the series 17, 23, 29….209. The number of terms in this
series of common terms would be 192/6 + 1 = 33. Option (d) is correct.
Most Upvoted Answer
Each of the series 13 + 15 + 17+…. and 14 + 17 + 20+… is c...
Solution:
To find the number of identical terms between the two series, we can first find the general term of each series and then equate them.
For the first series, the general term can be written as:
a1 = 13, a2 = 15, a3 = 17, ...
an = 2n + 11
For the second series, the general term can be written as:
b1 = 14, b2 = 17, b3 = 20, ...
bn = 3n + 11
Now, we need to find the values of n for which an = bn. Equating the two general terms, we get:
2n + 11 = 3n + 11
n = 0
This means that the first term of both series is identical. To find the next identical term, we can substitute n = 1 in the general terms:
a2 = 2(1) + 11 = 13
b2 = 3(1) + 11 = 14
These terms are not identical. We can continue this process until we find the first non-identical term. However, we can also notice that every third term of the first series is identical to every third term of the second series. This is because the difference between consecutive terms of the first series is 2, while the difference between consecutive terms of the second series is 3. Therefore, after every three terms, the difference between the two series increases by 1, and the two terms will no longer be identical.
Since both series have 100 terms, we can find the number of identical terms by dividing 100 by 3 and taking the floor of the result:
floor(100/3) = 33
Therefore, there are 33 identical terms between the two series, which is option (d).
To find the number of identical terms between the two series, we can first find the general term of each series and then equate them.
For the first series, the general term can be written as:
a1 = 13, a2 = 15, a3 = 17, ...
an = 2n + 11
For the second series, the general term can be written as:
b1 = 14, b2 = 17, b3 = 20, ...
bn = 3n + 11
Now, we need to find the values of n for which an = bn. Equating the two general terms, we get:
2n + 11 = 3n + 11
n = 0
This means that the first term of both series is identical. To find the next identical term, we can substitute n = 1 in the general terms:
a2 = 2(1) + 11 = 13
b2 = 3(1) + 11 = 14
These terms are not identical. We can continue this process until we find the first non-identical term. However, we can also notice that every third term of the first series is identical to every third term of the second series. This is because the difference between consecutive terms of the first series is 2, while the difference between consecutive terms of the second series is 3. Therefore, after every three terms, the difference between the two series increases by 1, and the two terms will no longer be identical.
Since both series have 100 terms, we can find the number of identical terms by dividing 100 by 3 and taking the floor of the result:
floor(100/3) = 33
Therefore, there are 33 identical terms between the two series, which is option (d).
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Each of the series 13 + 15 + 17+…. and 14 + 17 + 20+… is continued to 100 terms. Find howmany terms are identical between the two series.a)35b)34c)32d)33Correct answer is option 'D'. Can you explain this answer?
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