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13 May 2024, 23:46
Kudos
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Hi; we all member of family were stuck in the following,
Please kindly help and thanks a lot.
The sequence a, a+d, a+2d, a+3d, ……, a+(n-1)d has the following properties:
When the first, third and fifth, and so on terms are added, up to and including the last term, the sum is 320.
When the first, fourth, seventh, and so on, terms are added, up to and including the last term, the sum is 224.
What is the sum of the whole sequence?
Please kindly help and thanks a lot.
The sequence a, a+d, a+2d, a+3d, ……, a+(n-1)d has the following properties:
When the first, third and fifth, and so on terms are added, up to and including the last term, the sum is 320.
When the first, fourth, seventh, and so on, terms are added, up to and including the last term, the sum is 224.
What is the sum of the whole sequence?
14 May 2024, 23:30
Kudos
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Eric03
Let's denote the first term of the sequence as 𝑎 and the common difference as 𝑑.
Given that when the first, third, fifth, and so on terms are added, the sum is 320, we can express this sum as:
𝑆 1 = 𝑎 + ( 𝑎 + 2 𝑑 ) + ( 𝑎 + 4 𝑑 ) + … S 1 =a+(a+2d)+(a+4d)+…
The sum of this arithmetic series can be expressed using the formula for the sum of an arithmetic series:
𝑆 1 = 𝑛 2 [ 2 𝑎 + ( 𝑛 − 1 ) 𝑑 ] S 1 = 2 n [2a+(n−1)d] Where 𝑛 n is the number of terms.
Similarly, when the first, fourth, seventh, and so on terms are added, the sum is 224: 𝑆 2 = 𝑎 + ( 𝑎 + 3 𝑑 ) + ( 𝑎 + 6 𝑑 ) + … S 2 =a+(a+3d)+(a+6d)+…
We can again use the formula for the sum of an arithmetic series: 𝑆 2 = 𝑛 2 [ 2 𝑎 + ( 𝑛 − 1 ) 3 𝑑 ] S 2 = 2 n [2a+(n−1)3d]
Now, we have two equations: 𝑆 1 = 320 S 1 =320 𝑆 2 = 224 S 2 =224
Substituting the expressions for 𝑆 1 S 1 and 𝑆 2 S 2 into these equations,
we get: 320 = 𝑛 2 [ 2 𝑎 + ( 𝑛 − 1 ) 𝑑 ]
320= 2 n [2a+(n−1)d]
224 = 𝑛 2 [ 2 𝑎 + ( 𝑛 − 1 ) 3 𝑑 ]
224= 2 n [2a+(n−1)3d]
We can solve these two equations simultaneously to find the values of 𝑎 , 𝑑 , and 𝑛 .
However, there's an easier way to solve this problem.
Notice that in the second sequence, the terms are the multiples of 3 from the first sequence.
So, we can rewrite 𝑆 2 in terms of 𝑆 1 : 𝑆 2 = 𝑛 2 [ 2 𝑎 + ( 𝑛 − 1 ) 3 𝑑 ]
= 3 × 𝑛 2 [ 2 𝑎 + ( 𝑛 − 1 ) 𝑑 ]
= 3 𝑆 1 S 2 = 2 n [2a+(n−1)3d]=3× 2 n [2a+(n−1)d]=3S 1
So, we can conclude that 𝑆 2 is three times 𝑆 1.
Now, 𝑆 1 = 320, so 𝑆 2 = 3 × 320 = 960.
To find the sum of the whole sequence, we sum 𝑆 1 S 1 and 𝑆 2 S 2 :
Sum of the whole sequence = 𝑆 1 + 𝑆 2 = 320 + 960 = 1280
Sum of the whole sequence=S 1 +S 2 =320+960=1280
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