Find the nth term for the following pattern:11,21,31,41,51,61,............
also find it's 100th term
Answers
Answer:
an
Step-by-step explanation:
mark as brainliest answer
Answer:
the answer is
Step-by-step explanation:
General Form: an=a1+(n-1)d
an=1+(n-1)10
a1=1 (this is the 1st member)
an=71 (this is the last/nth member)
d=10 (this is the difference between consecutive members)
n=8 (this is the number of members)
Sum of finite series members
The sum of the members of a finite arithmetic progression is called an arithmetic series.
Using our example, consider the sum:
1+11+21+31+41+51+61+71
This sum can be found quickly by taking the number n of terms being added (here 8), multiplying by the sum of the first and last number in the progression (here 1 + 71 = 72), and dividing by 2:
n(a1+an)
2
8(1+71)
2
The sum of the 8 members of this series is 288
This series corresponds to the following straight line y=10x+1
Finding the nth element
a1 =a1+(n-1)*d =1+(1-1)*10 =1
a2 =a1+(n-1)*d =1+(2-1)*10 =11
a3 =a1+(n-1)*d =1+(3-1)*10 =21
a4 =a1+(n-1)*d =1+(4-1)*10 =31
a5 =a1+(n-1)*d =1+(5-1)*10 =41
a6 =a1+(n-1)*d =1+(6-1)*10 =51
a7 =a1+(n-1)*d =1+(7-1)*10 =61
a8 =a1+(n-1)*d =1+(8-1)*10 =71
a9 =a1+(n-1)*d =1+(9-1)*10 =81
a10 =a1+(n-1)*d =1+(10-1)*10 =91
a11 =a1+(n-1)*d =1+(11-1)*10 =101
a12 =a1+(n-1)*d =1+(12-1)*10 =111
a13 =a1+(n-1)*d =1+(13-1)*10 =121
a14 =a1+(n-1)*d =1+(14-1)*10 =131
a15 =a1+(n-1)*d =1+(15-1)*10 =141
a16 =a1+(n-1)*d =1+(16-1)*10 =151
a17 =a1+(n-1)*d =1+(17-1)*10 =161
a18 =a1+(n-1)*d =1+(18-1)*10 =171
a19 =a1+(n-1)*d =1+(19-1)*10 =181
a20 =a1+(n-1)*d =1+(20-1)*10 =191
a21 =a1+(n-1)*d =1+(21-1)*10 =201
a22 =a1+(n-1)*d =1+(22-1)*10 =211
a23 =a1+(n-1)*d =1+(23-1)*10 =221
a24 =a1+(n-1)*d =1+(24-1)*10 =231
a25 =a1+(n-1)*d =1+(25-1)*10 =241
a26 =a1+(n-1)*d =1+(26-1)*10 =251
a27 =a1+(n-1)*d =1+(27-1)*10 =261
a28 =a1+(n-1)*d =1+(28-1)*10 =271
a29 =a1+(n-1)*d =1+(29-1)*10 =281
a30 =a1+(n-1)*d =1+(30-1)*10 =291
a31 =a1+(n-1)*d =1+(31-1)*10 =301
a32 =a1+(n-1)*d =1+(32-1)*10 =311
a33 =a1+(n-1)*d =1+(33-1)*10 =321