Question

If the second term of an arithmetic sequence is 13. What is the sum of its 1st and 3rd terms?

Answer 1

Answer:

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

a1 = 10 and d = 3

Next term = a2=a1+d=10+3=13

a3=a2+d=13+3=16

Thus, 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

100=10+(n−1)3

90=(n−1)3

n−1=30

n=31, which is a whole number.

Therefore, 100 is the 31st term of the sequence.

Answer 2

Let a be the first term and d be the common difference of the APLet a be the first term and d be the common difference of the AP

a3=a+2d=13(1)(1)a3=a+2d=13

a9=a+8d=37(2)(2)a9=a+8d=37

by (2) - (1)by (2) - (1)

6d=24⟹d=46d=24⟹d=4

a+2d=13⟹a+2(4)=13⟹a+8=13⟹a=5a+2d=13⟹a+2(4)=13⟹a+8=13⟹a=5

Sum=n2(2a+(n−1)d)Sum=n2(2a+(n−1)d)

=62{2(5)+(6−1)(4)}=62{2(5)+(6−1)(4)}

=3(10+20)=3(30)=90=3(10+20)=3(30)=90