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the sequence is in AP and the second term would be 1st term plus the common difference. Now,2a +7d = a+6d => a+ d = 0 -> option B
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Official Solution




Given


    • In a sequence S, difference between any two terms is equal

      o S is an arithmetic sequence
      o Let the first term of the sequence be a and the common difference be d

    • 4th   term + 5th  term = 7th  term

      o 4th  term = a + (4-1)*d = a+ 3d
      o 5th  term = a + (5-1)*d = a +4d
      o 7th term = a + (7-1)d = a + 6d

To Find: a + d = ?


Working Out

    • (a + 3d) + (a+4d) = a+ 6d

      o 2a + 7d = a + 6d

      o a + d = 0

    • Hence, the value of the 2nd term of the sequence = a + d = 0



Answer: B

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a, a+d, a+2d, a+3d, a+4d, a+5d, a+6d, a+7d.

a4= a7-a5 = 2d

Therefore:
a2= a4 - d - d= 0

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In the sequence S, the difference between any two consecutive terms is equal. If the sum of the fourth term and the fifth term of the sequence is equal to the seventh term of the sequence, what is the value of the second term of the sequence?

    A. -4
    B. 0
    C. 4
    D. 8
    E. Cannot be Determined

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A very easy question!

Use concept of AP.

a+3d + a+4d = a+6d

a+d = 0
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a+3d+a+4d=a+6d [From problem statement]
a+d=0 -> This would be the 2nd term of the series.
Option B.
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EgmatQuantExpert
In the sequence S, the difference between any two consecutive terms is equal. If the sum of the fourth term and the fifth term of the sequence is equal to the seventh term of the sequence, what is the value of the second term of the sequence?

A. -4
B. 0
C. 4
D. 8
E. Cannot be Determined

let x+3=fourth term
x+4=fifth term
x+6=seventh term
adding, 2x+7=x+6➡
x=-1=first term
sequence=-1,0,1,2,3,4,5..
second term=0
B
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Deconstructing the Question

Since the difference between consecutive terms is constant, the sequence is arithmetic.

Let the first term be \(a\) and the common difference be \(d\).

Then the \(n\)th term is

\(T_n=a+(n-1)d\)

We are given that

\(T_4+T_5=T_7\)

and we need to find \(T_2\).

Step-by-step

Write the relevant terms:

\(T_4=a+3d\)

\(T_5=a+4d\)

\(T_7=a+6d\)

Use the condition:

\((a+3d)+(a+4d)=a+6d\)

Simplify:

\(2a+7d=a+6d\)

\(a+d=0\)

But

\(T_2=a+d\)

So

\(T_2=0\)

Answer: B
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