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Difficulty:
55%
(hard)
Question Stats:
66% (02:36) correct
34%
(02:09)
wrong
based on 164
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Suppose that {\(a_1\)} is an arithmetic sequence with \(a_1\) + \(a_2\) + . . . + \(a_{100}\) = 100 and \(a_{101}\) + \(a_{102}\)+ . . . + \(a_{200}\) = 200.
What is the value of \(a_2\) - \(a_1\)?
A. 0.0001
B. 0.001
C. 0.01
D. 0.1
E. 1
What is the value of \(a_2\) - \(a_1\)?
A. 0.0001
B. 0.001
C. 0.01
D. 0.1
E. 1
12 Apr 2022, 00:11
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I used the formula for the sum of Arithmetic Progression \(S=\frac{n}{2} [2a+(n-1).d]\). Where a is the first term of the AP, n is the number of terms and d is the common difference.
the given equations are -
\(a_1+a_2+.....+a_{100}=100\)---------------1 and
\(a_{101}+a_{102}+.....+a_{200}=200\)---------------2
adding these two, we get
\(a_1+a_2+.....+a_{100} + a_{101}+a_{102}+.....+a_{200}=300\)-----------3
now applying the formula of AP sum for APs 1 and 3 we get
\(100=\frac{100}{2} [2a_1+(100-1).(a_2-a_1)]\)
and
\(300=\frac{200}{2} [2a_1+(200-1).(a_2-a_1)]\)
these two equations can be easily solved for \((a_2 - a_1)=0.01\)
the given equations are -
\(a_1+a_2+.....+a_{100}=100\)---------------1 and
\(a_{101}+a_{102}+.....+a_{200}=200\)---------------2
adding these two, we get
\(a_1+a_2+.....+a_{100} + a_{101}+a_{102}+.....+a_{200}=300\)-----------3
now applying the formula of AP sum for APs 1 and 3 we get
\(100=\frac{100}{2} [2a_1+(100-1).(a_2-a_1)]\)
and
\(300=\frac{200}{2} [2a_1+(200-1).(a_2-a_1)]\)
these two equations can be easily solved for \((a_2 - a_1)=0.01\)
General Discussion
06 Nov 2024, 06:57
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧
Given that its an arithmetic series \(a_2\) - \(a_1\) = d = common difference
\(n^{th}\) term of the series is given by
\(T_n\) = \(a_1\) + (n-1)*d
=> \(T_{101}\) = \(a_{101}\) = \(a_1\) + (101-1)*d = \(a_1\) + 100d
=> \(T_{102}\) = \(a_{102}\) = \(a_1\) + (102-1)*d = \(a_1\) + d + 100d = \(a_2\) + 100d
=> \(T_{103}\) = \(a_{103}\) = \(a_3\) + 100d
.
.
.
=> \(T_{200}\) = \(a_{200}\) = \(a_{100}\) + 100d
=> \(a_{101}\) + \(a_{102}\)+ . . . + \(a_{200}\) = \(a_1\) + \(a_2\) + . . . + \(a_{100}\) + 100*100d
=> 200 = 100 + 10,000d
=> d = \(\frac{100}{10,000}\) = 0.01
So, Answer will be C
Hope it helps!
Watch the following video to MASTER Sequences
Given that its an arithmetic series \(a_2\) - \(a_1\) = d = common difference
\(n^{th}\) term of the series is given by
\(T_n\) = \(a_1\) + (n-1)*d
=> \(T_{101}\) = \(a_{101}\) = \(a_1\) + (101-1)*d = \(a_1\) + 100d
=> \(T_{102}\) = \(a_{102}\) = \(a_1\) + (102-1)*d = \(a_1\) + d + 100d = \(a_2\) + 100d
=> \(T_{103}\) = \(a_{103}\) = \(a_3\) + 100d
.
.
.
=> \(T_{200}\) = \(a_{200}\) = \(a_{100}\) + 100d
=> \(a_{101}\) + \(a_{102}\)+ . . . + \(a_{200}\) = \(a_1\) + \(a_2\) + . . . + \(a_{100}\) + 100*100d
=> 200 = 100 + 10,000d
=> d = \(\frac{100}{10,000}\) = 0.01
So, Answer will be C
Hope it helps!
Watch the following video to MASTER Sequences
