The sequence of four numbers \(a_1\), \(a_2\) , \(a_3\) and \(a_4\) is such that each number after the first is \(a_1-1\) greater than preceding number . What is the value of \(a_1\)?

Each number after the first is \(a_1-1\) greater than preceding number means that:
\(a_2=a_1+(a_1-1)=2a_1-1\)
\(a_3=a_2+(a_1-1)=3a_1-2\)
\(a_4=a_3+(a_1-1)=4a_1-3\)

(1) \(a_2=15\). We can find \(a_1\). Sufficient.

(2) \(a_4 = 29\). We can find \(a_1\). Sufficient.

Answer: D.

Hope it's clear.
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The sequence of four numbers \(a_1\), \(a_2\) , \(a_3\) and \(a_4\) is such that each number after the first is \(a_1-1\) greater than preceding number . What is the value of \(a_1\)?


\(a_n=a_{n-1}+(a_1-1)\)

1. \(a_2=15\)
\(a_2=a_{1}+(a_1-1)\)
\(15=2a_{1}-1\)
\(a_1=8\)
Sufficient

2. \(a_4 = 29\)
From the main formula above we can write \(a_4\) as \(a_4=a_{1}+3(a_1-1)\)
\(29=a_{1}+3(a_1-1)\)
\(29=4a_{1}-3\)
\(a_1=8\)
Sufficient

D

given:- a2 - a1 = a1 - 1 => a2 = 2a1 - 1

similarly we can get a3 = 3a1 - 2 and a4 = 4a1 - 3.

AD/BCE

statement 1:- a2 = 15 => 2a1 - 1 = 15.So we can get the value of a1. A alone is sufficient BCE out.

statement2:- a4 = 29 => 4a1-3 = 29 .So we can get the value of a1. B alone is sufficient.

Answer is D

From the given information, we know that \(a_n = a_{n-1} + (a_1 - 1) = na_1 - (n - 1)\)

1. \(a_2 = 15\)
--> \(a_2 = a_1 + (a_1 - 1) = 2a_1 - 1\) ----> \(2a_1 - 1 = 15\) ----> \(2a_1 = 16\) ----> \(a_1 = 8\)
Sufficient.

2. \(a_4 = 29\)
Now, \(a_4 = 4a_1 - 3\)
--> \(4a_1 - 3 = 29\) ----> \(4a_1 = 32\) ----> \(a_1 = 8\)
Sufficient.

Correct answer is D.

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Hope it helps.
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Bunuel please solve this question, I don't understand the solution for statement #2.

Thank you!

Given: \(a_1\), \(a_2\) , \(a_3\) and \(a_4\) and
\(a_2\) = \(a_1\) + \(a_1-1\) - (i)
\(a_3\) = \(a_1\) + 2\(a_1-1\) - (ii)
\(a_4\) = \(a_1\) + 3\(a_1-1\) - (iii)

Required: \(a_1\) = ?

Statement 1: \(a_2=15\)
Using (i) we can find the value of \(a_1\)
SUFFICIENT

Statement 2: \(a_4 = 29\)
Using (iii), we can find the value of \(a_1\)
SUFFICIENT

Hence Option D

Note: You do not need to solve for the values of \(a_1\). This can save precious time on the test.

Nth term in AP is denoted as tn=a+(n-1)d .

(1) says, 15=a1+(4-1)(a1-1) => 15=a1+3(a1-1) , from this a1 can be found. SUFFICIENT.

(2) says, 29=a1+(4-1)(a1-1) => 29=a1+3(a1-1) , from this a1 can be found. SUFFICIENT.

Thus answer D.
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Given: The sequence of four numbers a1, a2, a3 and a4 is such that each number after the first is a1 - 1 greater than preceding number
Let k = a1
So, each term after a1 is k - 1 greater than the term before it.

So we have:
a1 = k
a2 = k + (k - 1) = 2k - 1
a3 = 2k - 1 + (k - 1) = 3k - 2
a4 = 3k - 2 + (k - 1) = 4k - 3

Target question: What is the value of k?

Statement 1: a2 = 15
We already determined that a2 = 2k - 1
So, substitute 15 for a2 to get: 15 = 2k - 1
Solve: k = 8
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: a4 = 29
We already determined that a4 = 4k - 3
So, substitute 29 for a4 to get: 29 = 4k - 3
Solve: k = 8
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent
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