Question

2x+1, 4x-1, 5x+1.....are in arithmetic sequence
Find x
Write algebraic expression
Find the position of 195in this sequence

Answer 1

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✳ Required Answer:

✏ GiveN

• 2x + 1, 4x -1, 5x +1 are in AP

✏ To FinD:

• Find the value of x....?
• Find the position of 195 in the sequence.

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✳ How to solve?

For solving the above question we need to know the relation between the terms by it's common difference.

• Let three numbers a,b,c are in AP

Then, b - a = c - b = Common difference

➝ b - a = c - b

➝ 2b = a + c

✒ So, 2(second term) = First term + Third term

In this way, we can find x, After that we will get out first term and common difference. We have the nth term, So for finding which term it is, W can use....

$\large{ \boxed{ \rm{t_n = a + (n - 1)d}}}$

Where, tn is the nth term, a is first term, n is no. of terms.and d is the common difference.

✒ So, let's solve this question...

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✳ Solution:

Since, 2x + 1, 4x -1, 5x +1 are in AP

$\large{ \rm{ \therefore{2(4x - 1) = (2x + 1) + (5x + 1)}}} \\ \\ \large{ \rm{ \longrightarrow \: 8x - 2 = 7x + 2}} \\ \\ \large{ \rm{ \longrightarrow \: 8x - 7x = 2 + 2}} \\ \\ \large{ \rm{ \longrightarrow \: \boxed{ \red{ \rm{ x = 4}}}}}$

Now,

• First term = 2(4) + 1 = 9
• Common difference = Second term- first term

= 4x - 1 - (2x +1)

= 4x - 1 - 2x - 1

= 2x - 2

Putting the value of x

= 2(4) - 2 = 6

• nth term = 195

By using formula,

$\large{ \rm{ \longrightarrow \: 195 = 9 + (n - 1)6}} \\ \\ \large{ \rm{ \longrightarrow \: 186 = 6(n - 1)}} \\ \\ \large{ \rm{ \longrightarrow \: n - 1 = \cancel{\frac{186}{6} }}} \\ \\ \large{ \rm{ \longrightarrow \: n - 1 = 31}} \\ \\ \large{ \rm{ \longrightarrow \: \boxed{ \red{ \rm{n = 32}}}}}$

✏ Hence, 195 is the 32th term of the AP

$\large{ \therefore{ \underline{ \underline{ \purple{ \rm{Hence \: solved \: \dag}}}}}}$

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Answer 2

✪ Given ✪

2x + 1, 4x - 1, 5x + 1, ... are in arithmetic sequence.

✪ To Find ✪

• The value of x.
• The position of 195 in this sequence.

✪ Solution ✪

In an AP,

→ a₂ - a₁ = a₃ - a₂ = d

⇒4x - 1 - (2x + 1) = 5x + 1 - (4x - 1)

⇒4x - 1 - 2x - 1 = 5x + 1 - 4x + 1

⇒2x - 2 = x + 2

⇒2x - x = 2 + 2

⇒x = 4

$\rule{180}2$

By substituting the value of x in a₁ :-

2x + 1 = 2(4) + 1

⇒a₁ = 9

Common difference, d = a₂ - a₁

⇒d = 4(4) - 1 - a₁

⇒d = 15 - 9

⇒d = 6

$\rule{80}1$

Now,

Let 195 be the nth term of the AP.

We know,

aₙ = a + (n - 1)d

Substituting the values :-

195 = 9 + (n - 1)6

→195 - 9 = 6n - 6

→186 + 6 = 6n

→192 = 6n

→n = 192/6

→n = 32

Therefore, 195 is the 32nd term of the AP.