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:

$S=\frac{n}{2}\cdot\lbrack2a+(n-1)\cdot d\rbrack$

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.

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

Knowing that the sum is 1953:

$1953=\frac{n}{2}\cdot\lbrack2+4n\rbrack$

Multiplying both sides by 2:

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

Step 3: Use the Baskara formula to find n.

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

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

So, the equation:

$3906=2n+4n^2$

Can be written as:

$4n^2+2n-3906=0$

And, substituting it in the Bhaskara formula:

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

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

Answer:

There are 31 terms.