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Two consecutive terms in the expansion o...

Two consecutive terms in the expansion of `(3+2x)^74` have equal coefficients then term are (A) `30 and 31` (B) 38 and 39 (C) 31 and 32 (D) 37 and 38

A

`7^(th) and 8^(th)`

B

`11^(th) and 12(th)`

C

`30^(th) and 31^(th)`

D

`31^(th) and 32^(th)`

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To solve the problem of finding two consecutive terms in the expansion of \((3 + 2x)^{74}\) that have equal coefficients, we can follow these steps: ### Step 1: Identify the General Term The general term \(T_r\) in the expansion of \((a + b)^n\) is given by: \[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \] For our case, \(a = 3\), \(b = 2x\), and \(n = 74\). Therefore, the \(r\)-th term is: \[ T_{r+1} = \binom{74}{r} (3)^{74-r} (2x)^r \] ### Step 2: Write the Consecutive Terms The two consecutive terms are: - \(T_{r+1} = \binom{74}{r} (3)^{74-r} (2x)^r\) - \(T_{r+2} = \binom{74}{r+1} (3)^{74-(r+1)} (2x)^{r+1}\) ### Step 3: Set Up the Equation for Equal Coefficients We need to find \(r\) such that the coefficients of these two terms are equal: \[ \binom{74}{r} (3)^{74-r} (2^r) = \binom{74}{r+1} (3)^{73-r} (2^{r+1}) \] ### Step 4: Simplify the Equation Using the property of binomial coefficients, \(\binom{n}{r+1} = \frac{n-r}{r+1} \binom{n}{r}\), we can rewrite the equation: \[ \binom{74}{r} (3)^{74-r} (2^r) = \frac{74-r}{r+1} \binom{74}{r} (3)^{73-r} (2^{r+1}) \] Cancelling \(\binom{74}{r}\) from both sides (assuming \(r \neq 74\)), we get: \[ (3)^{74-r} (2^r) = \frac{74-r}{r+1} (3)^{73-r} (2^{r+1}) \] ### Step 5: Further Simplification Dividing both sides by \((3)^{73-r}\) and \(2^r\): \[ 3 = \frac{74-r}{r+1} \cdot 2 \] This simplifies to: \[ 3(r+1) = 2(74-r) \] ### Step 6: Solve for \(r\) Expanding and rearranging gives: \[ 3r + 3 = 148 - 2r \] \[ 5r = 145 \implies r = 29 \] ### Step 7: Find the Terms Now, we need to find the two consecutive terms: - \(T_{30} = T_{r+1} = T_{29+1}\) - \(T_{31} = T_{r+2} = T_{29+2}\) Thus, the terms are \(T_{30}\) and \(T_{31}\). ### Conclusion The two consecutive terms in the expansion of \((3 + 2x)^{74}\) that have equal coefficients are: \[ \text{Terms: } 30 \text{ and } 31 \] Thus, the answer is (A) 30 and 31. ---
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AAKASH INSTITUTE ENGLISH-BINOMIAL THEOREM-Assignment (section-A)
  1. The middle terms in the expansion of (1+x)^(2n+1) is (are)

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  2. (1.003)^(4) is nearby equal to

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  3. The nubmber of non - zeroes terns in the expansion of (1+sqrt(5))^(6)+...

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  4. The number of non -zeroes terms in the expansion of (sqrt(7)+1)^(75)-(...

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  5. The number of terms in the expansion if (a+b+c)^(12) is

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  6. Two consecutive terms in the expansion of (3+2x)^74 have equal coeffic...

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  7. If the coefficients of rth, (r+ 1)th and (r + 2)th terms in the expa...

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  8. Cofficient of x^(3)y^(10)z^(5) in expansion of (xy+yz+zx)^(6) is

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  9. The ratio of coefficients x^(3) and x^(4) in the expansion of (1+x)^(1...

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  10. Given the integers r gt 1, n gt 2 and coefficients of (3r) th and (r+2...

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  11. Find the coefficient of x^5 in the expansion of (1+x^2)^5(1+x)^4.

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  12. If (r+1)^(th) term in the expasnion of (a^(3)/3-2/a^(2))^(10) contains...

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  13. Find n and x in the expansion of (1 + x)^n, if the fifth term is four ...

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  14. Cofficients of x^(6)y^(3) in the expansion of (x+y)^(9) is

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  15. The number of terms in the expansion of (4x^(2) + 9y^(2) + 12xy)^(6) ...

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  16. The middle term in the expansioin of (2x-1/3x)^(10) is

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  17. The coefficient of the term independent of x in the expansion of (a x+...

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  18. Find the middle term in the expansion of (x- 1/(2x))^12

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  19. The value of .^(13)C(7)+.^(13)C(8)+.^(13)C(9)+.^(13)C(10)+.^(13)C(11)+...

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  20. For all natural number of n, 2^(2n).3^(2n)-1-35n is divisible by

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