Question

The sum of first 4 terms of an arithmetic sequence is 120 and the first term is 18 a)What is the common difference b)Write the sequence​

Answer 1

Answer:

First term of the sequence is 10 and common difference is 3.

First term of the sequence is 10 and common difference is 3.a1 = 10 and d = 3

First term of the sequence is 10 and common difference is 3.a1 = 10 and d = 3Next term = a2=a1+d=10+3=13

First term of the sequence is 10 and common difference is 3.a1 = 10 and d = 3Next term = a2=a1+d=10+3=13a3=a2+d=13+3=16

First term of the sequence is 10 and common difference is 3.a1 = 10 and d = 3Next term = a2=a1+d=10+3=13a3=a2+d=13+3=16Thus, first three terms of the sequence are 10, 13 and 16.

First term of the sequence is 10 and common difference is 3.a1 = 10 and d = 3Next term = a2=a1+d=10+3=13a3=a2+d=13+3=16Thus, first three terms of the sequence are 10, 13 and 16.Let 100 be the nth term of the sequence.

First term of the sequence is 10 and common difference is 3.a1 = 10 and d = 3Next term = a2=a1+d=10+3=13a3=a2+d=13+3=16Thus, first three terms of the sequence are 10, 13 and 16.Let 100 be the nth term of the sequence.an=a1+(n−1)d

First term of the sequence is 10 and common difference is 3.a1 = 10 and d = 3Next term = a2=a1+d=10+3=13a3=a2+d=13+3=16Thus, first three terms of the sequence are 10, 13 and 16.Let 100 be the nth term of the sequence.an=a1+(n−1)d100=10+(n−1)3

First term of the sequence is 10 and common difference is 3.a1 = 10 and d = 3Next term = a2=a1+d=10+3=13a3=a2+d=13+3=16Thus, first three terms of the sequence are 10, 13 and 16.Let 100 be the nth term of the sequence.an=a1+(n−1)d100=10+(n−1)390=(n−1)3

First term of the sequence is 10 and common difference is 3.a1 = 10 and d = 3Next term = a2=a1+d=10+3=13a3=a2+d=13+3=16Thus, first three terms of the sequence are 10, 13 and 16.Let 100 be the nth term of the sequence.an=a1+(n−1)d100=10+(n−1)390=(n−1)3n−1=30

First term of the sequence is 10 and common difference is 3.a1 = 10 and d = 3Next term = a2=a1+d=10+3=13a3=a2+d=13+3=16Thus, first three terms of the sequence are 10, 13 and 16.Let 100 be the nth term of the sequence.an=a1+(n−1)d100=10+(n−1)390=(n−1)3n−1=30n=31, which is a whole number.

First term of the sequence is 10 and common difference is 3.a1 = 10 and d = 3Next term = a2=a1+d=10+3=13a3=a2+d=13+3=16Thus, first three terms of the sequence are 10, 13 and 16.Let 100 be the nth term of the sequence.an=a1+(n−1)d100=10+(n−1)390=(n−1)3n−1=30n=31, which is a whole number.Therefore, 100 is the 31st term of the sequence.

Step-by-step explanation:

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