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The sequence \(a_1\), \(a_2\), \(a_3\) ... \(a_n\) is defined such that \(a_n=a_{n-1}+9+n\) for all \(n>1\). If \(a_1=10\), what is the value of \(a_{11}\)?
(A) 150
(B) 155
(C) 160
(D) 165
(E) 170
(A) 150
(B) 155
(C) 160
(D) 165
(E) 170
27 Apr 2020, 13:49
Kudos
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\(a_2 = a_1 + 9+2 = a_1+11\)
\(a_3 = a_2+9+3= a_1+11+12\)
\(a_{11}= a_1+11+12+....+(9+11)\)
\(a_{11} = 10+11+12....+20 = \frac{(20+10)}{2} * 11 = 165\)
\(a_3 = a_2+9+3= a_1+11+12\)
\(a_{11}= a_1+11+12+....+(9+11)\)
\(a_{11} = 10+11+12....+20 = \frac{(20+10)}{2} * 11 = 165\)
Ravixxx
KhulanE
24 Feb 2021, 19:25
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\(A_n\)=\(A_{n-1}\)+9+n we can convert this formula into this one. -> \(A_n\)-\(A_{n-1}\)=9+n
now we can substitute numbers n to numbers from 1 to 11 just like below one
\(A_{11}\)-\(A_{10}\)=9+11
\(A_{10}\)-\(A_9\)=9+10
\(A_9\)-\(A_8\)=9+9
...
\(A_3\)-\(A_2\)=9+3
\(A_2\)-\(A_1\)=9+2
if you add up above all, all left is below equation
\(A_{11}\)-\(A_1\)=9*10+(2+3+...+11)=90+\(\frac{11*12}{2}\)-1=155
\(A_{11}\)-10=155
\(A_{11}\)=165
if you wonder how this part (2+3+...+11) ended up to this part (\(\frac{11*12}{2}\)-1), refer to the below explanation.
2+3+...+11=(1+2+3+...+11)-1=\(\frac{11*12}{2}\)-1=65
here is the formula of the consecutive integers from 1 to n: sum=\(\frac{n*(n+1)}{2}\)
now we can substitute numbers n to numbers from 1 to 11 just like below one
\(A_{11}\)-\(A_{10}\)=9+11
\(A_{10}\)-\(A_9\)=9+10
\(A_9\)-\(A_8\)=9+9
...
\(A_3\)-\(A_2\)=9+3
\(A_2\)-\(A_1\)=9+2
if you add up above all, all left is below equation
\(A_{11}\)-\(A_1\)=9*10+(2+3+...+11)=90+\(\frac{11*12}{2}\)-1=155
\(A_{11}\)-10=155
\(A_{11}\)=165
if you wonder how this part (2+3+...+11) ended up to this part (\(\frac{11*12}{2}\)-1), refer to the below explanation.
2+3+...+11=(1+2+3+...+11)-1=\(\frac{11*12}{2}\)-1=65
here is the formula of the consecutive integers from 1 to n: sum=\(\frac{n*(n+1)}{2}\)
