Question

is 132 a term of the series 6,11,16 ?
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Answer 1

Explanation :-

The given sequence of numbers is an arithmetic sequence ( A.P.) . AP is a sequence in which common difference between two consecutive terms is always constant throughout every term. If we subtract any succeeding term from it's preceding term we will get the common difference of AP. There are certain general equations in the concept of AP, which will help us to solve this problem.

Solution :-

Given sequence of numbers is 6, 11, 16 . . .

First term of AP, represented by a is 6.

Common difference of AP, represented by d is 5. ( 11 - 6 ) or ( 16 - 11 ).

General form of nth term of AP is ::

• $\sf A_n = A + ( n - 1 ) d$

Here,

• $\sf A_n=n^{th}\; term$
• $\sf n \:= \:Number\:of\:terms$
• $\sf a \; = \; First\:term$
• $\sf d \: = \: Common\; difference.$

Firstly we are assuming that 132 is a term of AP. Let's say it's kth term. So,

$\sf\implies\; A_n \;= \;a\:+\:(\:n\:-\:1\:)\:d$

$\sf\implies\; 132 \;= \;6\:+\:(\:k\:-\:1\:)\:5$

$\sf\implies\; 132 \;-6\:=\:5k\:-\:5$

$\sf\implies\; 126\:=\:5k\:-\:5$

$\sf\implies\; 126\:+5\:=\:5k$

$\sf\implies\; 131\:=\:5k$

$\sf\implies\; \dfrac{131}{5}\:=\:5k$

Now since value of k is not an integer, this means that 132 is not a term of AP because any term of AP can be indexed with it's integer position such as 4th term, 8th term 19th term etc.

Hence 132 is not a term of AP.