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Bunuel
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In an sequence of numbers a_1, a_2, a_3, ..., the difference between s 14 May 2020, 01:51
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B
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In an sequence of numbers \(a_1\), \(a_2\), \(a_3\), ..., the difference between successive terms is constant. If \(a_3 + a_9 = 8\), what is the sum of the first 11 terms of the sequence?

A. 55
B. 44
C. 33
D. 22
E. 11
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B
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In an sequence of numbers a_1, a_2, a_3, ..., the difference between s 14 May 2020, 02:26
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\(a_3=a_1+2d\), where d is the difference between successive terms

\(a_9=a_1+8d \)

\(a_3+a_9 = a_1+2d+a_1+8d = 2a_1+10d = 8\)

\(2*(a_1+5d)=8\)

\((a_1+5d)=4\)

Sum of first 11 terms= Mean*number of terms

Mean of the first 11 terms = Median of first 11 terms = 6th term = \(a_1+5d\)

Sum of first 11 terms= \((a_1+5d)*11\) = \(4*11 = 44\)
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In an sequence of numbers a_1, a_2, a_3, ..., the difference between s 14 May 2020, 02:35
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Difference is common so it becomes an AP.

a3 + a9 = 8

a1+ 2 *d + a1 + 8*d = 8

2*a1 + 10 *d = 8

So find sum of 1st 11 terms --> 11/2(2 * a1 + 10 * d) --> 11/2*8 = 44

Ans is 44 (B)
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In an sequence of numbers a_1, a_2, a_3, ..., the difference between s 14 May 2020, 04:11
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The given sequence is an arithmetic sequence since the difference between successive terms is constant.
In an arithmetic sequence \(a_1\), \(a_2\), \(a_3\)….. \(a_n\), \(a_1\) + \(a_n\) = \(a_2\) + \(a_{n-1}\) = \(a_3\) + \(a_{n-2}\) and so on.

If there are 11 terms in the sequence, \(a_1\)+ \(a_{11}\) = \(a_2\) + \(a_{10}\) = \(a_3\) + \(a_9\) and so on.

But, \(a_3\) + \(a_9\) = 8. Therefore, \(a_1\) + \(a_{11}\) = 8.

In an arithmetic sequence, Arithmetic Mean = \(\frac{First Term + Last Term }{ 2}\).

Therefore, in the given sequence, AM = \(\frac{a_1+ a_{11} }{ 2}\) = 8 / 2 = 4.

But, Arithmetic Mean =\(\frac{ Sum of terms }{ Number of terms}\). Therefore,

Sum of terms = Arithmetic Mean * Number of terms = 4 * 11 = 44.
The correct answer option is B.

Hope that helps!
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In an sequence of numbers a_1, a_2, a_3, ..., the difference between s 14 May 2020, 04:30
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If a1 = a and difference between successive term is b,
a3 + a9 = 8
(a + 2b) + (a + 8b) = 8
2a + 10b = 8

Sum of the first 11 terms of the sequence
= n/2 * (a1 + an)
= 11/2 * (a + a + 10b)
= 11/2 * (2a+10b)
= 11/2 * 8
= 44

FINAL ANSWER IS (B)

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In an sequence of numbers a_1, a_2, a_3, ..., the difference between s 14 May 2020, 04:55
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Quote:
In an sequence of numbers a1a1, a2a2, a3a3, ..., the difference between successive terms is constant. If a3+a9=8, what is the sum of the first 11 terms of the sequence?

A. 55
B. 44
C. 33
D. 22
E. 11
Show more

a1=x
a2=x+d(n-1)=x+d
a3=8-a9, x+2d=8-(x+8d),
2x+10d=8, [a11=x+10d],
x+10d=8-x

sum is average * n
a1+a11/2*11=x+(8-x)/2*11=44

ans (B)
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In an sequence of numbers a_1, a_2, a_3, ..., the difference between s 14 May 2020, 05:25
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I think the option A should be 54.
The sequence starting from -1 to 9
Sum is 54.

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In an sequence of numbers a_1, a_2, a_3, ..., the difference between s 14 May 2020, 05:53
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In an sequence of numbers \(a_1, a_2, a_3\), ..., the difference between successive terms is constant. If \(a_3+a_9=8\), what is the sum of the first 11 terms of the sequence?

A. 55
B. 44
C. 33
D. 22
E. 11

\(a_ 2 - a_1 = a_3 - a_2\)
\(a_3 = a_1 + 2d\) & \(a_9 = a_3 + 6d = a_1 + 8d\)

\(a_3+a_9 = 8\)
\(a_1 + 2d + a_1 + 8d = 8\)
\(2a_1 + 10d = 8\)
\(a_1 + 5d = 4\)
\(a_6 = 4\)

Since sequence has odd numbers of terms i.e. 11, mean = middle term = \(a_6\)
Sum of sequence of 11 terms = Mean of the sum of sequence * Number of terms = 4 * 11 = 44

Answer B.
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In an sequence of numbers a_1, a_2, a_3, ..., the difference between s 14 May 2020, 20:26
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In an sequence of numbers a1, a2, a3, ..., the difference between successive terms is constant. If a3+a9=8, what is the sum of the first 11 terms of the sequence?

A. 55
B. 44
C. 33
D. 22
E. 11
n/2[2a+(n-1)d]
11/2[2a+10d]
we have 2a+10d=8,
4[11]
44
Ans B
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In an sequence of numbers a_1, a_2, a_3, ..., the difference between s 14 May 2020, 22:34
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Concept - 1

If all the numbers are in A.P (as indicated in question successive terms) then Average of all the symmetrical combination are SAME .

Here a3 and a9 are in symmetrical (as they are 3rd position from left and right)

So, Average is a middle term a6

a6 = (a3 + a9) / 2 -- > 8/2 = 4

Concept -2

Middle term of on odd-AP is 4 so 5 numbers are in left and 5 numbers are in right side which will get cancel , why , here it is

d = common difference .

a1, a2, a3, a4,a5,a6,a7,a8,a9,a10,a11 are now as

4 - 5d , 4 - 4d , 4 - 3d , 4 - 2d , 4 - d , 4 , 4 + d , 4 + 2d , 4 + 3d , 4 + 4d , 4 +5d

All d's will be cancel out …. = remaining ---> 4 * 11 = 44


Ans - 44 , Option B - Correct
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In an sequence of numbers a_1, a_2, a_3, ..., the difference between s 17 Mar 2025, 18:23
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