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In the sequence a1, a2, a3 ,…, an, an = an−1 + 1. If a1=b then (a10)^2

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In the sequence a1, a2, a3 ,…, an, an = an−1 + 1. If a1=b then (a10)^2 [#permalink] New post   15 Dec 2020, 02:29
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In the sequence \(a_{1}, a_{2}, a_{3}\),…, \(a_{n}\), \(a_{n}=a_{n−1}+1.\) If \(a_{1}=b\) then \((a_{10})^2–(a_{7})^2\) is equal to which of the following?

A. \(3b\)

B. \(6b+45\)

C. \(6b+51\)

D. \(7b+45\)

E. \(7b+51\)
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B
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Re: In the sequence a1, a2, a3 ,…, an, an = an−1 + 1. If a1=b then (a10)^2 [#permalink] New post   15 Dec 2020, 03:36
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Sajjad1994 wrote:
In the sequence \(a_{1}, a_{2}, a_{3}\),…, \(a_{n}\), \(a_{n}=a_{n−1}+1.\) If \(a_{1}=b\) then \((a_{10})^2–(a_{7})^2\) is equal to which of the following?

A. \(3b\)

B. \(6b+45\)

C. \(6b+51\)

D. \(7b+45\)

E. \(7b+51\)




Given a1 = b and \(a_n = a_{n-1} + 1\)

a2 = a1 + 1 = b + 1

a3 = a2 + 1 = (b + 1) + 1 = b + 2

a4 = (b + 2) + 1 = b + 3

So \(a_n = b + (n - 1)\)

a10 = b + 9 and a7 = b + 6

\((a_{10} - a_7)^2 = (a_{10} + a_7)(a_{10} - a_7)\)

= (b + 9 + b + 6)(b + 9 - b - 6) = (2b + 15)(3) = 6b + 45


Option B

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Re: In the sequence a1, a2, a3 ,…, an, an = an−1 + 1. If a1=b then (a10)^2 [#permalink] New post   15 Dec 2020, 03:58
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Sajjad1994 wrote:
In the sequence \(a_{1}, a_{2}, a_{3}\),…, \(a_{n}\), \(a_{n}=a_{n−1}+1.\) If \(a_{1}=b\) then \((a_{10})^2–(a_{7})^2\) is equal to which of the following?

A. \(3b\)

B. \(6b+45\)

C. \(6b+51\)

D. \(7b+45\)

E. \(7b+51\)


Since \(a_{n}=a_{n−1}+1\) and \(a_{1}=b\)
\(a_{n} - a_{n−1} = 1\) &
\(a_{2}=b+1\)
\(a_{3}=b+2\)
.
.
.
\(a_{7}=b+6\)

\((a_{10})^2–(a_{7})^2\) = \((a_{10} – a_{7})*(a_{10} + a_{7})\)
\((a_{10} – a_{9} + a_{9} - a_{8} + a_{8} - a_{7})*(a_{10} - a_{9} + a_{9} - a_{8} + a_{8} + a_{7})\)
\((1 + 1 + 1)*(1 + 1 + a_{7} + 1 + a_{7})\)
\(3*(3 + 2a_{7} )\)
\(3*(3 + 2(b + 6))\)
\(3*(2b + 15)\)
\((6b + 45)\)

Answer B
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In the sequence a1, a2, a3 ,…, an, an = an−1 + 1. If a1=b then (a10)^2 [#permalink] New post   25 Dec 2020, 08:10
Sajjad1994 wrote:
In the sequence \(a_{1}, a_{2}, a_{3}\),…, \(a_{n}\), \(a_{n}=a_{n−1}+1.\) If \(a_{1}=b\) then \((a_{10})^2–(a_{7})^2\) is equal to which of the following?

A. \(3b\)

B. \(6b+45\)

C. \(6b+51\)

D. \(7b+45\)

E. \(7b+51\)


Alternate Soln:

Take \(b= 1 \) then \(a_{1}, a_{2},a_{3},a_{3},a_{4}...a_{7}..a_{10}= 1,2,3,4,...7..10\)

So \(a_{10}=10 \) and \( a_{7}= 7\)

\((a_{10})^2 - ( a_{7})^2= 100-49 = 51,\) Check for \(b=1\) which option gives 51. Only option B works.

Ans-B

Hope it's clear.
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In the sequence a1, a2, a3 ,…, an, an = an−1 + 1. If a1=b then (a10)^2 [#permalink]
25 Dec 2020, 08:10
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