Question

the algebra of an arithmetic progression is 3n+ 1 how could we say that the squares of the terms of the sequence are in that sequence itself by looking the algebra. ( AP-4,7,10)​

Answer 1

Answer:

Hint: We find the common difference using terms of the given sequence. Form algebraic form of AP using general form of nth term of AP. Put nth term as 100 and check if it satisfies the equation of nth term of AP. For part (c) we take the square of the general term obtained in part (a) and use the factorization method to reduce it into smaller forms.

Complete step-by-step answer:

(a)

We are given the arithmetic sequence 1, 4, 7, 10…

Here first term is 1

⇒a=1

Since second term is 4, we can find the common difference by subtracting first term from second term

⇒d=4−1

⇒d=3

We know any arithmetic sequence has nth term given by the formula an=a+(n−1)d

Here we substitute the value of a=1,d=3

⇒an=1+(n−1)3

Multiplying the terms in RHS

⇒an=1+3n−3

⇒an=3n−1

∴

Algebraic form of arithmetic series is an=3n−2

(b)

We have to check if the number 100 is a term of the AP

Substitute the value of an=100

in equation (1)

⇒100=3n−2

Shift all constant values to one side of the equation

⇒3n=100+2

⇒3n=102

Divide both sides by 3

⇒3n3=1023

Cancel same factors from numerator and denominator

⇒n=34