Consider the sequence of numbers which gives the remainder 2 on diving by 3. a) write the sequence. b) what is the algebraic from of the sequence? c) Is 101 a term of this sequence?
Answers
Given that:Consider the sequence of numbers which leaves the remainder 3 on dividing by 4.
a) write the sequence.
b) write the algebraic form of this sequence.
c) which is the term just below 100
d) how many terms are there below 100 in this sequence.
e) calculate the sum of all terms below 100
Solution:
As on divided by 4 number leaves remainder 3,
So,
First number is 7
Next number will be 7+4= 11
This way all the terms will be generated
a) write the sequence.
Ans: 7,11,15,19,23...99
b) write the algebraic form of this sequence.
Ans: Algebraic form : 4n+3
n= 0,1,2,...
c) which is the term just below 100
Ans: 99
d) how many terms are there below 100 in this sequence.
Ans: It generates an A.P.
here a(First term)=7
d(common difference)=4
Last term(l)=99
To find number of terms,apply this formula
\begin{gathered}\bold{a_n = a + (n - 1)d} \\ \\ 99 = 7 + (n - 1)4 \\ \\ 99 - 7 = 4(n - 1) \\ \\ 92 = 4(n - 1) \\ \\ (n - 1) = \frac{92}{4} \\ \\ n - 1 = 23 \\ \\ n = 24 \\ \\ \end{gathered}an=a+(n−1)d99=7+(n−1)499−7=4(n−1)92=4(n−1)(n−1)=492n−1=23n=24
Total terms= 24
e) calculate the sum of all terms below 100
Ans:
Sum of terms can be find by applying formula of sum of n terms
\begin{gathered}\bold{S_n = \frac{n}{2} (a + l)} \\ \\ Sum = \frac{24}{2} (7 + 99) \\ \\ Sum = 12 \times 106 \\ \\ Sum = 1272 \\ \\ \end{gathered}Sn=2n(a+l)Sum=224(7+99)Sum=12×106Sum=1272
Sum of terms below 100 are 1272.
Hope it helps you.