Question $3 / 7$
Find the sum of the 1 st 112 terms of: $7+11$ $+15+\ldots .$

Studdy Solution

STEP 1

1. The sequence given is an arithmetic sequence.
2. The first term of the sequence is $a = 7$ .
3. The common difference $d$ between consecutive terms is $4$ (since $11 - 7 = 4$ ).
4. We need to find the sum of the first 112 terms of this sequence.

STEP 2

1. Identify the formula for the sum of an arithmetic sequence.
2. Calculate the last term of the sequence.
3. Use the formula to find the sum of the first 112 terms.

STEP 3

The formula for the sum $S_n$ of the first $n$ terms of an arithmetic sequence is:
$S_n = \frac{n}{2} \times (a + l)$
where $n$ is the number of terms, $a$ is the first term, and $l$ is the last term.

STEP 4

Calculate the last term $l$ of the sequence using the formula for the $n$ -th term of an arithmetic sequence:
$l = a + (n-1) \times d$
Given $a = 7$ , $d = 4$ , and $n = 112$ , we find:
$l = 7 + (112-1) \times 4$ $l = 7 + 111 \times 4$ $l = 7 + 444$ $l = 451$

STEP 5

Substitute the values into the sum formula to find the sum of the first 112 terms:
$S_{112} = \frac{112}{2} \times (7 + 451)$ $S_{112} = 56 \times 458$
Calculate the product:
$S_{112} = 56 \times 458 = 25,648$
The sum of the first 112 terms of the sequence is:
$\boxed{25,648}$

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