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4) how many terms of the sequence 3, 7, 11, 15, …..yield a sum of 1953?

4) how many terms of the sequence 3, 7, 11, 15, …..yield a sum of 1953?-example-1
User Allan Ho
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1 Answer

4 votes

Given:

a = first term

n = number of terms

d = commom difference

For an arithmetic sequence, the sum (S) of the terms is:


S=(n)/(2)\cdot\lbrack2a+(n-1)\cdot d\rbrack

To solve this question, you have to find "n". To do it, follow the steps.

Step 1: Find "a" and "d"

a is the first term of the sequence, that is, a = 3.

d is the common difference and can be found by subtracting two consecutive numbers. d = 7 - 3; d = 4.

Step 2: Substitute the values in the equation.


\begin{gathered} S=(n)/(2)\cdot\lbrack2a+(n-1)\cdot d\rbrack \\ S=(n)/(2)\cdot\lbrack2\cdot3+(n-1)\cdot4\rbrack \\ S=(n)/(2)\cdot\lbrack6+(n-1)\cdot4\rbrack \\ S=(n)/(2)\cdot\lbrack6+4n-4\rbrack \\ S=(n)/(2)\cdot\lbrack2+4n\rbrack \\ \end{gathered}

Knowing that the sum is 1953:


1953=(n)/(2)\cdot\lbrack2+4n\rbrack

Multiplying both sides by 2:


\begin{gathered} 1953\cdot2=(n)/(2)\cdot(2+4n)\cdot2 \\ 3906=n\cdot(2+4n) \\ 3906=2n+4n^2 \end{gathered}

Step 3: Use the Baskara formula to find n.

For a quadratic equation ax2 + bx + c = 0, the roots x are:


x=(-b\pm√(b^2-4ac))/(2a)

So, the equation:


3906=2n+4n^2

Can be written as:


4n^2+2n-3906=0

And, substituting it in the Bhaskara formula:


\begin{gathered} n=\frac{-2\pm\sqrt[]{2^2-4\cdot4\cdot(-3906)}}{2\cdot4} \\ n=\frac{-2\pm\sqrt[]{^{}4+62496}}{8} \\ n=\frac{-2\pm\sqrt[]{^{}62500}}{8} \\ n=(-2\pm250)/(8) \\ n_1=(-2+250)/(8)=(248)/(8)=31 \\ n_2=(-2-250)/(8)=(-252)/(8)=-31.5 \end{gathered}

Since the number of terms is a positive number, n = 31.

Answer:

There are 31 terms.

User Gonzih
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