Question

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The sum of a sequence of consecutive integers is 342. The largest integer in the sequence is 3 times greater than the smallest integer in the sequence. What is the smallest integer and how many integers are in the sequence?

Answer 1

Answer:

smallest integer=9,

total number of integers are in the sequence  = 19.

Step-by-step explanation:

• let a be the first term and the number of terms be n. as they are consecutive,the common difference will be 1. these will form Arithmetic Progression series.

• the sum of n terms in AP series is : $\frac{n}{2} (a+l)$

[ where, l: last term]

• or,  sum of n terms in AP series is :$\frac{n}{2}(2a+[n-1]d)$

[where, d: common difference]

using the first formula,

342 = $\frac{n}{2} (a+3a)$

342= $\frac{n}{2} (4a)$

342=2an  .......equation 1.

using the second formula,

342=$\frac{n}{2} (2a+{n-1}l)$

684=n[2a+n-1]

684=2an+n^2-n

from equation 1,

342=n^2-n   .......equation 1

$n^{2}$-n-342=0

after solving, the solution to the above equation is 19.

substituting this in equation 2,

171=a[19]

a=9

so, smallest integer= a = 9,

total number of integers are in the sequence = n = 19.

Answer 2

Answer:

Smallest integer is 9

number of integers is 19

Step-by-step explanation:

Answer:

Step-by-step explanation:

Sum on consecutive integers is

n(a+l)/2

where a is the first/smallest integer, l is the last/largest integer and n is the number of integers

n(a+l)/2 = 342 (1)

l = 3a (2)

l = a + (n-1) (3)

Using (2) and (3)

3a = a+(n-1)

2a = n-1

n = 2a+1 (4)

Using (1) (2) and (4)

(2a+1)(a+3a)/2 = 342

(2a+1)(4a) = 684

4(2a²+a) = 684

2a²+a= 684÷4

2a²+a = 171

2a²+a-171 = 0

2a²+19a-18a-171 = 0

a(2a+19)-9(2a+19) = 0

(2a+19)(a-9) = 0

a = 9 or a = -19/2

Since a is an integer, it has to be 9

a = 9

n = 2a+1 = 2(9)+1

n = 19