Question

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

Answer 1

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.