The sum of all the terms common to the arithmetic progressions 1, 3, 5, ....., 1991 and 1, 6, 11, ......., 1991, is? (1) 199100 (2) 199200 (3) 199300 (4) 200196

Answer 1

To find the sum of all the terms common to the arithmetic progressions 1, 3, 5, ..., 1991 and 1, 6, 11, ..., 1991, we need to first find the last term that appears in both progressions.

For the first progression (1, 3, 5, ..., 1991), the last term is 1991. For the second progression (1, 6, 11, ..., 1991), we need to find the last term that is less than or equal to 1991. This can be calculated using the formula for the nth term of an arithmetic progression: (a_n = a_1 + (n-1)d), where (a_n) is the nth term, (a_1) is the first term, (n) is the number of terms, and (d) is the common difference.

For the second progression, (a_1 = 1) and (d = 6 - 1 = 5). We need to find (n) such that (a_1 + (n-1)d \leq 1991). [ 1 + (n-1)5 \leq 1991 ] [ (n-1)5 \leq 1990 ] [ n-1 \leq \frac{1990}{5} ] [ n-1 \leq 398 ] [ n \leq 399 ]

So, the last term of the second progression that is less than or equal to 1991 is the 399th term. Now, we can find the sum of the common terms using the formula for the sum of an arithmetic progression: (S = \frac{n}{2}(a_1 + a_n)).

For the first progression, (a_1 = 1), (a_n = 1991), and (n = \frac{1991 - 1}{2} + 1 = 996). [ S_1 = \frac{996}{2}(1 + 1991) = 996 \times 996 = 992016 ]

For the second progression, (a_1 = 1), (a_n = 1991), and (n = 399). [ S_2 = \frac{399}{2}(1 + 1991) = 399 \times 1992 = 794808 ]

Therefore, the sum of all the terms common to the arithmetic progressions is (S_1 + S_2 = 992016 + 794808 = 1786824).

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Answer 2

(2) #199200#

Given:

#1, 3, 5,...,1991#
#1, 6, 11,...,1991#
Note that the common difference of the first sequence is #2# and that of the second is #5#.
Since these have no common factor greater than #1#, their least common multiple is #10#, which is the common difference of the intersection of the two sequences:
#1, 11, 21, 31,..., 1991#
This sequence has #200# terms, with average value:
#1/2 * (1+1991) = 1992/2#

So the sum is:

#200*1992/2 = 199200#
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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