Question

if five times the 3rd team is equal to three times the 5th term of an AP then prove that 3rd term is divisible by 3​

Answer 1

Answer:

5(a+2d)=3(a+4d).

[OPEN THE PARENTHESES AND SOLVE]

2a=2d

a=d

3rd Term

==>a+2d

Since a=d,

==>a+2a

3a

Which Is Divisible By 3

Answer 2

Answer:

Proof:

3rd term of AP, $T_{3} = a+(3-1)d = a+2d$

5th term of AP, $T_{5} = a+(5-1)d = a+4d$

Given that, 5 times the 3rd term is equal to 3 times the 5th term of AP. That is,

$5 (a+2d) = 3(a+4d)$

Opening the brackets,

⇒ $5a+10d = 3a+12d$

⇒ $5a - 3a = 12d-10d$

⇒ $2a = 2d$

⇒ $a=d$

We got $a=d$. Substituting this in the equation of 3rd term,

$T_{3} = a+2d$

⇒ $T_{3} = a+2a$    (∵ $a=d$ )

⇒ $T_{3} = 3a$

That is,  $T_{3}$ is divisible by 3.

Step-by-step explanation:

Arithmetic Progression:

An Arithmetic sequence or progression is defined as a sequence of numbers in which the difference between two consecutive terms is always a fixed constant.

The nth term of an Arithmetic Progression (AP) is defined by the formula,

$T_{n} = a+(n-1)d$

Learn more about Arithmetic Progression:

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