Question

If (a +1), 3a and (4a +2) be
any three
Consecutive terms of an AP, fine the
value of a​

Answer 1

Answer:

a=3

Step-by-step explanation:

((a+1)+(4a+2))/2=3a

5a+3=6a

a=3

Answer 2

$\huge\sf\pink{Answer}$

☞ Your Answer is 3

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$\huge\sf\blue{Given}$

✭ (a+1) , 3a and (4a + 1) is in A.P

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$\huge\sf\gray{To \:Find}$

◈ The Value of a?

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$\huge\sf\purple{Steps}$

The $\sf n^{th}$ term of an AP is given by,

$\underline{\boxed{\sf a_n = a + (n-1)d}}$

Finding The Common difference

➝ $\sf d = 3a - (a + 1)$

➝ $\sf \red{d = 2a - 1}$

Finding The Value of a(check the 3rd term)

»» $\sf 3a + (3-1)(2a - 1) = 4a + 2$

»» $\sf a+1 + (2)(2a - 1) = 4a + 2$

»» $\sf a+1 + 4a - 2 = 4a + 2$

»» $\sf 5a - 1 = 4a + 2$

»» $\sf\orange{ a = 3}$

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