# Q: Give the sequence 2, 6, 12, 20, 30, ... 110, k, 156,..., Find k.________

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`a_(n+1)=11^2+11=132`

`a_n-a_(n-1)=2n`

Since `a_1=2` `a_1-a_0=2 xx 1` thus `a_0=0`

Then the sequence given `S_n` is a sum of the sequence:

`0,2,4,6,8,10.........2n`

So that `S_n=` `sum_(k=0)^n 2k =2 sum_(k=0)^n k= 2 xx (n^2+n)/2=n^2+n`

Now for `S_n=110` :

Let it be `a_(n+1)` the anknow term.

`a_(n+1)-a_n= a_(n+1)-110=2(n+1)`

and:

`a_(n+2)-a_(n+1)=156 - a_(n+1)=2(n+2)`

then:

`156-110=a_(n+2)-a_n=a_(n+2)-a_(n+1) + a_(n+1)- a_n=`

`=2(n+2)+2(n+1)=4n+6`

From trhe relation: `46=4n+6` we get: `n=10`

Now we can easly find the missing term:

``

Integer point on blue line are the sqeuence `a_n=2n` , while the integers points on red line are the sum sequence `S_n=n^2+n` ``

`a_(n+1)-a_n=2xxn+2`

`a_2-a_1=4`

`a_3-a_2=6`

`a_(n+1)=a_n+2(n+1)`

` ` `k=110+2(m-1+1)`

`k=110+2m` (i)

156=k+2(m+1) (ii)

156=K+2+k-110

156=2k-108

264=2k

k=132

Ans.

Here we have assume let m th term is k. then simplify